Question

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

Answer 1

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

Explanation:

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!

Answer 2

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

Explanation:

`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`