# Multiple trig integral

1. Sep 25, 2014

### mshiddensecret

1. The problem statement, all variables and given/known data

Integrate: (sin(2x))^3(cos2x)^2dx

2. Relevant equations

Using substitution

Cos2x= (1-(sinx)^2)

3. The attempt at a solution

I sub u= sin2x

But then got nowhere because I had cos2x to the power of 2 and I don't know how to compensate for it with du.

2. Sep 26, 2014

### haruspex

First, an easy substitution gets rid of the 2 factors in the 2x terms.
When you have a mix of sin and cos in an integral dx, look for combining one of them with the dx, e.g. cos(x)dx = d sin(x).
In the present case, you can choose a cos or a sin. Which works better?

3. Sep 26, 2014

### HallsofIvy

Staff Emeritus
When you have even powers of both sine and cosine, reduce the powers using $sin^2(x)= (1/2)(1- cos(2x))$ and $cos^2(x)= (1/2)(1+ cos(2x))$.

4. Sep 26, 2014

### pasmith

I can't tell whether you mean $\cos^2 x= 1 - \sin^2 x$, which is true, or $\cos 2x = 1 - \sin^2 x$, which is false: $\cos 2x = \cos^2 x - \sin^2 x = 1 - 2\sin^2 x$.

You have an extra power of $\sin 2x = -\frac12 \frac{d}{dx} \cos 2x$. That suggests $u = \cos 2x$, not $u = \sin 2x$.

5. Sep 26, 2014

### haruspex

It's simpler than that. See my post #2.

6. Sep 26, 2014

### mshiddensecret

I got it. You use he identity and make sin into 1-cos. and then sub u for cos 2x and work from there.

7. Sep 28, 2014

### HallsofIvy

Staff Emeritus