How do you determine if the improper integral converges or diverges #int ln(sin(x))# from 0 to pi/2?

After using Feynman's trick,

After integrating both sides,

ToTo determine if the improper integral ∫ ln(sin(x)) fromTo determine if the improper integral ∫ ln(sin(x)) from To determine if the improper integral ∫ ln(sin(x)) from 0To determine if the improper integral ∫ ln(sin(x)) from 0 toTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2To determine if the improper integral ∫lnTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converTo determine if the improper integral ∫ln(sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 convergesTo determine if the improper integral ∫ln(sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverTo determine if the improper integral ∫ln(sin(x))To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or divergesTo determine if the improper integral ∫ln(sin(x)) fromTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges,To determine if the improper integral ∫ln(sin(x)) from 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, weTo determine if the improper integral ∫ln(sin(x)) from 0 toTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'llTo determine if the improper integral ∫ln(sin(x)) from 0 to πTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyzeTo determine if the improper integral ∫ln(sin(x)) from 0 to π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior ofTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 convergesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of lnTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges orTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sinTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x))To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges,To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) nearTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, weTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we canTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundariesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyzeTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries ofTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze theTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behaviorTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the intervalTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior ofTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of lnTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1.To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. NearTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) asTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

2. Near xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

3. Near x =To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approachesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

4. Near x = To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

5. Near x = 0To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

6. Near x = 0: To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 andTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

7. Near x = 0: -To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and πTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

8. Near x = 0:

   • lnTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

9. Near x = 0:

   • ln(sinTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

10. Near x = 0:

    • ln(sin(xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:
   • ln(sin(x))To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:
   • ln(sin(x)) approaches negativeTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1.To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity asTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

2. Near x =To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

3. Near x = 0:

   • ln(sin(x)) approaches negative infinity as xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

4. Near x = To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

5. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approachesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

6. Near x = 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

7. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

8. Near x = 0: To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

9. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 fromTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

10. Near x = 0: To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

11. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

12. Near x = 0: AsTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

13. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the rightTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

14. Near x = 0: As x approachesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

15. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

16. Near x = 0: As x approaches To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

17. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (sinceTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

18. Near x = 0: As x approaches 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

19. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sinTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

20. Near x = 0: As x approaches 0,To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

21. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

22. Near x = 0: As x approaches 0, sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

23. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x)To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

24. Near x = 0: As x approaches 0, sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

25. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approachesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

26. Near x = 0: As x approaches 0, sin(x)To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

27. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0). To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

28. Near x = 0: As x approaches 0, sin(x) approaches 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

29. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0). To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

30. Near x = 0: As x approaches 0, sin(x) approaches 0,To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

31. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0). -To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

32. Near x = 0: As x approaches 0, sin(x) approaches 0, andTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

33. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • ThisTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

34. Near x = 0: As x approaches 0, sin(x) approaches 0, and lnTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

35. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggestsTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

36. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

37. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests aTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

38. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

39. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergenceTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

40. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approachesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

41. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence atTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

42. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negativeTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

43. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

44. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinityTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

45. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x =To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

46. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity.To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

47. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

48. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. ThisTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

49. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggestsTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

2. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potentialTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

2. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2.To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergenceTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

2. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

3. NearTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

4. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence atTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

5. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

6. Near xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

7. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

8. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

9. Near x =To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

10. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x =To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

11. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

12. Near x = πTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

13. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

14. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

15. Near x = π/2To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

16. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2: To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

3. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2: -To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

3. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

4. NearTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

5. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

6. Near x = π/2:

   • lnTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

7. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

8. Near xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

9. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

10. Near x = π/2:

    • ln(sinTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

11. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

12. Near x =To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

13. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

14. Near x = π/2:

    • ln(sin(xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

15. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

16. Near x = πTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

17. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

18. Near x = π/2:

    • ln(sin(x))To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

19. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

20. Near x = π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

21. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

22. Near x = π/2:

    • ln(sin(x)) approachesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

23. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

24. Near x = π/2To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

25. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

26. Near x = π/2:

    • ln(sin(x)) approaches negativeTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

27. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

28. Near x = π/2: To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

29. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

30. Near x = π/2:

    • ln(sin(x)) approaches negative infinityTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

31. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

32. Near x = π/2: To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

33. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

34. Near x = π/2:

    • ln(sin(x)) approaches negative infinity asTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

35. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

36. Near x = π/2: AsTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

37. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

38. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

39. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

40. Near x = π/2: As xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

41. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

42. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approachesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

43. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

44. Near x = π/2: As x approachesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

45. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

46. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches πTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

47. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

48. Near x = π/2: As x approaches πTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

49. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

50. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

51. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

52. Near x = π/2: As x approaches π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

53. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

54. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 fromTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

55. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

56. Near x = π/2: As x approaches π/2,To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

57. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

58. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

59. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

60. Near x = π/2: As x approaches π/2, sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

61. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

62. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the leftTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

63. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

64. Near x = π/2: As x approaches π/2, sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

65. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

66. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

67. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

68. Near x = π/2: As x approaches π/2, sin(x)To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

69. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

70. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (sinceTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

71. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

72. Near x = π/2: As x approaches π/2, sin(x) approachesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

73. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

74. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sinTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

75. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

76. Near x = π/2: As x approaches π/2, sin(x) approaches To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

77. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

78. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

79. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

80. Near x = π/2: As x approaches π/2, sin(x) approaches 1To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

81. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

82. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x)To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

83. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

84. Near x = π/2: As x approaches π/2, sin(x) approaches 1,To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

85. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

86. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approachesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

87. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

88. Near x = π/2: As x approaches π/2, sin(x) approaches 1, andTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

89. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

90. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

91. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

92. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and lnTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

93. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

94. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

95. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

96. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

97. Near x = 0:

    • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
    • This suggests a potential divergence at x = 0.

98. Near x = π/2:

    • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1). To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

99. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

100. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

101. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

102. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1). To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

103. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

104. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x))To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

105. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

106. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1). -To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

107. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

108. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approachesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

109. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

110. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • ThisTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

111. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

112. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

113. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

114. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggestsTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

115. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

116. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

117. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

118. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests anotherTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

119. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

120. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0.To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

121. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

122. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potentialTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

123. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

124. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. ThisTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

125. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

126. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergenceTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

127. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

128. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggestsTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

129. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

130. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence atTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

131. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

132. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergenceTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

133. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

134. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence at xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

135. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

136. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence atTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

137. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

138. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence at x =To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

139. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

140. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

141. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

142. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence at x = πTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

143. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

144. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x =To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

145. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

146. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence at x = π/To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

147. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

148. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = πTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

149. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

150. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence at x = π/2To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

151. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

152. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

153. Near x = 0:

     • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
     • This suggests a potential divergence at x = 0.

154. Near x = π/2:

     • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
     • This suggests another potential divergence at x = π/2.

To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

3. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

4. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

SinceTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since lnTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

ThereforeTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sinTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore,To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(xTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, theTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x))To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integralTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tendsTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends towardTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫lnTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negativeTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinityTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity atTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x))To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at bothTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) fromTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpointsTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints ofTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 toTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the intervalTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to πTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval,To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, theTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improperTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integralTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 convergesTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integral diverTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converges sinceTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integral divergesTo determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converges since lnTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integral diverges.To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converges since ln(sinTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integral diverges.To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converges since ln(sin(xTo determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integral diverges.To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converges since ln(sin(x))To determine if the improper integral ∫ ln(sin(x)) from 0 to π/2 converges or diverges, we'll analyze the behavior of ln(sin(x)) near the boundaries of the interval.

1. Near x = 0:

   • ln(sin(x)) approaches negative infinity as x approaches 0 from the right (since sin(x) approaches 0).
   • This suggests a potential divergence at x = 0.

2. Near x = π/2:

   • ln(sin(x)) approaches negative infinity as x approaches π/2 from the left (since sin(x) approaches 1).
   • This suggests another potential divergence at x = π/2.

Since ln(sin(x)) tends toward negative infinity at both endpoints of the interval, the improper integral diverges.To determine if the improper integral ∫ln(sin(x)) from 0 to π/2 converges or diverges, we can analyze the behavior of ln(sin(x)) as x approaches 0 and π/2.

1. Near x = 0: As x approaches 0, sin(x) approaches 0, and ln(sin(x)) approaches negative infinity. This suggests a potential divergence at x = 0.

2. Near x = π/2: As x approaches π/2, sin(x) approaches 1, and ln(sin(x)) approaches 0. This suggests convergence at x = π/2.

Therefore, the integral ∫ln(sin(x)) from 0 to π/2 converges since ln(sin(x)) approaches a finite limit as x approaches π/2.