How do you integrate #int x^3 sqrt(-x^2 - 16x-39)dx# using trigonometric substitution?

Question

Elijah Abram · Accepted Answer

#int x^3 sqrt(-x^2-16x-39) dx = 25 int (5sint+8)^3( 1-sin^2t)dt#

Write:

#-x^2-16x-39 = -x^2 -16x -64+25 = 25 -(x-8)^2#

so that we get a difference of squares under the root.

Now substitute:

#x-8 = 5sint#

#dx = 5costdt#

As the function is defined only for:

#abs(x-8) <= 5#

we have #t in (-pi/2,pi/2)#

Performing the substitution we get:

#int x^3 sqrt(-x^2-16x-39) dx = 5 int (5sint+8)^3 sqrt (25-25sin^2t) costdt = 25 int (5sint+8)^3 cos^2tdt = 25 int (5sint+8)^3( 1-sin^2t)dt#

where in #cost =+- sqrt(1-sin^2t) # we took the plus sign as for #t in (-pi/2,pi/2)# the cosine is positive.

We have now to integrate a polynomial in #sint#, which can be easily calculated using reduction formulas.

Charles Arocha · Answer

undefined To integrate $ \int x^3 \sqrt{-x^2 - 16x - 39} \, dx $ using trigonometric substitution, perform the following steps:

1. Complete the square under the square root:
   $ -x^2 - 16x - 39 = -(x^2 + 16x + 64) + 25 = -((x + 8)^2 - 25) = -(x + 8)^2 + 25 $.

2. Let $ x + 8 = 5\sec(\theta) $.
   Then, $ dx = 5\sec(\theta)\tan(\theta) \, d\theta $.

3. Substitute into the integral:
   $ \int (5\sec(\theta) - 8)^3 \sqrt{-(5\sec(\theta))^2} \cdot 5\sec(\theta)\tan(\theta) \, d\theta $.

4. Simplify the expression:
   $ \int (5\sec(\theta) - 8)^3 \cdot 5\tan(\theta) \cdot 5\sec(\theta)\tan(\theta) \, d\theta $.

5. Use trigonometric identities to simplify:
   $ \int (5\sec(\theta) - 8)^3 \cdot 25\sec(\theta)\tan^2(\theta) \, d\theta $.

6. Let $ u = 5\sec(\theta) - 8 $.
   Then, $ du = 5\sec(\theta)\tan(\theta) \, d\theta $.

7. Substitute $ u $ and $ du $ into the integral:
   $ \int u^3 \cdot 25\sec(\theta)\tan^2(\theta) \, d\theta $.

8. Rewrite $ \sec^2(\theta) $ in terms of $ u $:
   $ \sec^2(\theta) = \frac{u^2}{25} + 1 $.

9. Substitute $ \sec^2(\theta) $ and $ du $ in terms of $ u $ into the integral:
   $ \int u^3 \cdot \frac{25u^2}{25} \, du $.

10. Simplify and integrate:
   $ \int u^3 \cdot u^2 \, du = \int u^5 \, du $.

11. Integrate $ u^5 $:
   $ \frac{u^6}{6} + C $.

12. Substitute back for $ u $:
   $ \frac{(5\sec(\theta) - 8)^6}{6} + C $.

13. Finally, substitute back for $ \theta $ in terms of $ x $:
   $ \frac{(5\sec(\arccos(\frac{x + 8}{5})) - 8)^6}{6} + C $.