How do you find the integral of #int (6sin^6x)(cos^3x)dx#?

Question

Luke Alvarez · Accepted Answer

#I = (6sin^7x)/7-(2sin^9x)/3+C# #I = int (6sin^6x)(cos^3x)dx = 6int sin^6x cos^2x cosxdx# #I = 6int sin^6x (1-sin^2x) cosxdx# #sinx=t => cosxdx=dt# #I = 6int t^6 (1-t^2) dt = 6 int (t^6-t^8)dt = 6 (t^7/7-t^9/9+C)# #I = (6sin^7x)/7-(2sin^9x)/3+C#

William Abdalla · Answer

undefined To find the integral of $ \int (6\sin^6(x))(\cos^3(x)) \, dx $, we can use trigonometric identities and integration techniques.

Let $ u = \sin(x) $ and $ dv = \cos^3(x) \, dx $. Then, $ du = \cos(x) \, dx $ and $ v = \frac{1}{4}\sin^4(x) $. Applying integration by parts:

$$ \int (6\sin^6(x))(\cos^3(x)) \, dx = \frac{1}{4}(6\sin^6(x))(\sin^4(x)) - \frac{1}{4}\int (6\sin^4(x))(\sin^4(x)) \, dx $$

Now, we need to integrate $ \int (6\sin^4(x))(\sin^4(x)) \, dx $. We can use the power-reducing formula $ \sin^2(x) = 1 - \cos^2(x) $ to express $ \sin^4(x) $ in terms of $ \cos^2(x) $.

$$ \sin^4(x) = (\sin^2(x))^2 = (1 - \cos^2(x))^2 $$

Expanding this, we get $ 1 - 2\cos^2(x) + \cos^4(x) $. Therefore,

$$ \int (6\sin^4(x))(\sin^4(x)) \, dx = \int (6\sin^4(x))(1 - 2\cos^2(x) + \cos^4(x)) \, dx $$

Now, we can integrate each term separately. After integrating, we get the final result for the integral $ \int (6\sin^6(x))(\cos^3(x)) \, dx $.