How do you determine if the improper integral converges or diverges #int ln(sin(x))# from 0 to pi/2?
This integral converges to
After using Feynman's trick,
After integrating both sides,
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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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.
 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.
 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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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