How do you evaluate the integral #int cos^3xsin^3x#?

Question

Ezra Adams · Accepted Answer

#=1/4sin^4x - 1/6sin^6x + C#

We want to rewrite so that either sine or cosine is all by itself with a power of #1# and use a u-substitution to eliminate it, leaving all terms in u that are easier to integrate.

#=int cosx(cos^2xsin^3x)#

Rewrite #cos^2x# using the pythagorean identity #sin^2x + cos^2x = 1#.

#=int cosx(1 - sin^2x)sin^3xdx#

#=int cosx(sin^3x - sin^5x)dx#

The derivative of #sinx# is #cosx#. Accordingly, we take the substitution #u = sinx#. Then #du = cosxdx# and #dx = (du)/cosx#

#=int cosx(u^3 - u^5) * (du)/cosx#

This will eliminate, leaving us only with u's.

#=int u^3 - u^5 du#

Integrate using #x^ndx = x^(n + 1)/(n + 1) + C#.

#=1/4u^4 - 1/6u^6 + C#

Resubstitute:

#=1/4(sinx)^4 - 1/6(sinx)^6 + C#

#=1/4sin^4x - 1/6sin^6x + C#

Hopefully this helps!

Emily Ammerman · Answer

The answer is #=1/192cos6x-3/64cos2x+C#

#cos^3x=1/4(cos3x+3cosx)#

#sin^3x=1/4(3sinx-sin3x)#

#cos^3x*sin^3x=1/16(cos3x+3cosx)(3sinx-sin3x)#

#=1/16(3cos3xsinx-cos3xsin3x+9cosxsinx-3sin3xcosx)#

#9cosxsinx=9/2sin2x#

#cos3xsin3x=1/2sin6x#

#-3sin3xcosx+3cos3xsinx=-3(sin(3x-x))=-3sin2x#

Therefore,

#cos^3x*sin^3x=1/16(9/2sin2x-3sin2x-1/2sin6x)#

#=3/32sin2x-1/32sin6x#

So,

#intcos^3x*sin^3xdx=3/32intsin2xdx-1/32intsin6xdx#

#=(3/32*-cos2x*1/2)-(1/32*-cos6x*1/6)+C#

#=1/192cos6x-3/64cos2x+C#

Christopher Agresta · Answer

undefined To evaluate the integral $ \int \cos^3(x) \sin^3(x) $, you can use a trigonometric identity to simplify the integrand. Here's how:

1. Use the identity $ \sin(2x) = 2\sin(x)\cos(x) $ to rewrite the integrand:
   $ \sin^3(x) \cos^3(x) = (\sin(x)\cos(x))^3 = \left(\frac{\sin(2x)}{2}\right)^3 $
2. Expand the cube:
   $ \left(\frac{\sin(2x)}{2}\right)^3 = \frac{\sin^3(2x)}{8} $
3. Now, use a substitution to simplify the integral:
   Let $ u = 2x $, then $ du = 2dx $ or $ dx = \frac{du}{2} $
   The integral becomes:
   $ \frac{1}{8} \int \sin^3(u) du $
4. Use another trigonometric identity, $ \sin^3(u) = \frac{3\sin(u) - \sin(3u)}{4} $:
   $ \frac{1}{8} \int \left(\frac{3\sin(u) - \sin(3u)}{4}\right) du $
5. Integrate each term separately:
   $ \frac{1}{8} \left(\frac{3}{4}\int \sin(u) du - \frac{1}{4} \int \sin(3u) du\right) $
   $ \frac{1}{8} \left(\frac{3}{4}(-\cos(u)) - \frac{1}{4}\left(-\frac{\cos(3u)}{3}\right)\right) $
6. Simplify the expression:
   $ \frac{1}{8} \left(\frac{-3\cos(u)}{4} + \frac{\cos(3u)}{12}\right) $
   $ \frac{-3\cos(u)}{32} + \frac{\cos(3u)}{96} $
7. Substitute back $ u = 2x $:
   $ \frac{-3\cos(2x)}{32} + \frac{\cos(6x)}{96} $

So, the integral $ \int \cos^3(x) \sin^3(x) dx $ simplifies to $ \frac{-3\cos(2x)}{32} + \frac{\cos(6x)}{96} + C $, where $ C $ is the constant of integration.