Question

Question #6dbbf

Answer 1

`int\ cot^2(2x)sin^3(2x)\ dx=-1/6cos^3(2x)+C`

Explanation:

I will use the following identity to rewrite the expression:
`cot(theta)=cos(theta)/sin(theta)`

In the case of our integral, this gives:
`int\ cos^2(2x)/cancel(sin^2(2x))sin^cancel(3)(2x)\ dx=`

`=int\ cos^2(2x)sin(2x)\ dx`

To solve this, I will introduce a u-substitution with `u=cos(2x)`. We can using the chain rule get that the derivative will be:
`(du)/dx=-2sin(2x)`

This means that to integrate with respect to `u`, we divide through by `-2sin(2x)`:
`int\ (cos^2(2x)cancel(sin(2x)))/(-2cancel(sin(2x)))\ du=`

`int\ -1/2u^2\ du=-1/2int\ u^2\ du`

This integral is a straight application of the reverse power rule:
`-1/2*u^3/3+C`

Undoing the substitution, we get that our answer is:
`-1/6cos^3(2x)+C`

Answer 2

`-1/6cos^3(2x)+C`

Explanation:

Since `cot(2x) = cos(2x)/sin(2x)` we can rewrite the integrand:

`cot^2(2x)sin^3(2x) = cos^2(2x)/sin^2(2x)*sin^3(2x)`

`=cos^2(2x)sin(2x)`

So the integral can be rewritten:

`int cos^2(2x)sin(2x)dx`

We can do this by u-substitution.

Let `u=cos(2x)` so `du=-2sin(2x)dx`

Rearranging our `du` we have:

`-1/2du=sin(2x)dx`

Now we can make substitutions to get:

`-1/2int u^2du = -1/2(1/3)u^3+C`

Simplifying and substituting for `u`:

`int cot^2(2x)sin^3(2x)dx = -1/6cos^3(2x)+C`