# Integration with trig

1. Feb 3, 2014

### jdawg

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

∫8sin4x dx

2. Relevant equations

3. The attempt at a solution
8∫sin2xsin2x

I'm not really sure what to do next. Maybe substitute a 1-cos2x in for one of the sin2x? Maybe both?

2. Feb 3, 2014

### Dick

That sort of integration involves using double angle identities in trig. Look them up and try and get started.

3. Feb 3, 2014

### ehild

Do you know the relations

$$\sin^2(x)=\frac{1-\cos(2x)}{2}$$ and $$\cos^2(x)=\frac{1+\cos(2x)}{2}$$

ehild

4. Feb 3, 2014

### jdawg

Ok, so I plugged in (1-cos(2x))/2 for both of my sin2x and then foiled.
This is what I have now:
2∫1+cos22x+2cos2x dx

5. Feb 3, 2014

### Dick

No, that's not what you get. There's a sign problem. But the important point is that you can use the double angle identities on $\cos^2(2x)$ as well.

Last edited: Feb 3, 2014
6. Feb 3, 2014

### jdawg

Oh, so it should be: 2∫1+cos22x-2cos2x dx
Then: 2∫cos22x-cos2x dx
Now could you split up your integrals and solve them individually?

7. Feb 4, 2014

### Dick

Of course you can split them up and solve them individually. That's the whole point. But now what happened to the "1" part? And a factor of 2 also disappeared. Just take it step by step.

Last edited: Feb 4, 2014