# Thread: Trouble with Trig integral

1. ## Trouble with Trig integral

For some reason I am having trouble with the integral: sin2xcos2x dx
I have tried a few identities with no luck.

1st try
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(1/2) | (2sinxcosx)2 dx

2nd try
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(1/4) | (1 - cos(2x))(1 + cos(2x)) dx

The website I am looking at keeps saying to use integration by parts, which makes no sense to me.

2. ## Re: Trouble with Trig integral

first step : Integral sin^2x cos^2x dx = 1/4 Integral sin^2 ( 2x ) dx
Now use the identity cos 2x = 1 - sin^2 x

3. ## Re: Trouble with Trig integral

Hello,

$\displaystyle \sin^2x\cos^2x = (\sin x \cos x)^2$

You can use the double angle formula:

$\displaystyle (\sin x \cos x) = \frac{\sin (2x)}{2}$

Thus

$\displaystyle (\sin x \cos x)^2 = \frac{\sin^2(2x)}{4}$

Now write $\displaystyle \sin^2(2x) = 1 - \cos^2(2x)$ and subsequently $\displaystyle \cos^2(2x) = \frac{1}{2} + \frac{\cos (4x)}{2}$. So, $\displaystyle \sin^2(2x) = 1 - (\frac{1}{2} + \frac{\cos (4x)}{2}) = \frac{1}{2} - \frac{\cos (4x)}{2}$.

So, the integral is now: $\displaystyle \frac{1}{4}(\frac{1}{2} - \frac{\cos (4x)}{2}) = \frac{1}{8} - \frac{\cos (4x)}{8}$.

Integrate that and you get $\displaystyle \frac{x}{8} - \frac{\sin (4x)}{32}$. The only identities you need are $\displaystyle \sin^2x + \cos^2x = 1$ and the double angle formulae $\displaystyle \sin (2x) = 2\sin x\cos x$, and $\displaystyle \cos (2x) = \cos^2x - \sin^2x$. No integration by parts is necessary.

4. ## Re: Trouble with Trig integral

I think you're along the right lines for one possible solution with your first try.

Using the identity:

$\displaystyle \sin2x = 2\sin x\cos x$

we can say:

$\displaystyle \int \sin^2x\cos^2x \ dx = \int (\sin x\cos x)^2 \ dx = \frac{1}{4}\int (2\sin x\cos x)^2 \ dx = \frac{1}{4}\int \sin^2 2x \ dx$

now, in my understanding we can't integrate $\displaystyle \sin^2x$, so:

$\displaystyle \cos 4x = \cos^2 2x - \sin^2 2x$

$\displaystyle \cos 4x = 1 - 2\sin^2 2x$

$\displaystyle \sin^2 2x = \frac{1-\cos 4x}{2}$

therefore the integral now becomes:

$\displaystyle \frac{1}{8} \int (1 - \cos 4x) \ dx = \frac{1}{8}[x - \frac{1}{4} \sin 4x] = \frac{1}{8}x - \frac{1}{32}\sin 4x + c$

5. ## Re: Trouble with Trig integral

Qwob, you missed that the angle is doubled twice so you end up with $\displaystyle \sin(4x)$, and of course the coefficient then halves.

6. ## Re: Trouble with Trig integral

Thanks Naranja, fixed it now.

7. ## Re: Trouble with Trig integral

It looks like it was a little more in depth than I thought, thanks guys! =)

8. ## Re: Trouble with Trig integral

This link gives some quick advice about how to proceed with combined sine/cosine integrals. It doesn't actually explain why these rules of thumb work, but if you do several examples I think you'll figure out the why's by yourself. (Or we can help, too!)

-Dan