# Can't seem to figure this integral out

I have an integral here that I'm trying to figure out.

$$\int 7\sin^2x \cos^4x\ dx$$

Here's what I got as an answer:

$$\frac{7}{16}x-\frac{7}{64}\sin4x+\frac{7}{12}\sin2x + C$$

However, I'm doubting myself and the check didn't seem to produce good results. I can give some steps if you want. I filled a whole page with work, and it seems like it should be easier than that. Any ideas?

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wolframalpha.com/input/?i=int+7sin%28x%29%5E2+cos%28x%29%5E4 Click on "show steps" for one possible derivation. – Bill Cook Oct 10 '11 at 0:41
@BillCook: That uses the reduction formula though. I'd rather stick to identities and such. – confused Oct 10 '11 at 0:43

You can use $\sin(x)\cos(x) = {1 \over 2}\sin(2x)$ and $\cos^2(x) = {1 + \cos(2x) \over 2}$ and your integral is $$7 \int\left({1 \over 2}\sin(2x)\right)^2 {1 + \cos(2x) \over 2}\,dx$$ $$= {7 \over 8}\int \sin^2(2x)\,dx + {7 \over 8}\int\sin^2(2x)\cos(2x)\,dx$$ The first term is just a $\sin^2$ integral, and the second can be dealt with by a $u$ substitution $u = \sin^2(2x)$.

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That looks easy. I actually did the first step that you suggested, but somehow I made it a lot harder than it needed to be. Thanks for the help! – confused Oct 10 '11 at 1:35
Hmm, I did it again when I had the chance and I got this: $$\frac{7}{16}x-\frac{7}{64}\sin4x+\frac{7}{12}\sin^32x + C$$ Unfortunately, that doesn't seem to check according to WolframAlpha. – confused Oct 10 '11 at 14:34
@confused I think the coefficient before the $\sin^3(2x)$ term should be ${7 \over 48}$. In other words your $du$ should be $2 \cos(2x)\,dx$ and not $2 du = \cos(2x)\,dx$. – Zarrax Oct 10 '11 at 21:44
Yep, I just found that out. I had taken the antiderivative of $u$ instead of the derivative. However, even with that coefficient it still doesn't check. – confused Oct 10 '11 at 21:49
There are multiple ways of writing a given function sometimes, so most likely they are the same function but in different forms. – Zarrax Oct 10 '11 at 22:05

Hint: Try writing $\sin^2(x)=1-\cos^2(x)$ and then use the identity $\cos^2(x)=\frac{1+\cos(2x)}{2}$.

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...some algebra...apply identities again...some algebra...pain and suffering. – Bill Cook Oct 10 '11 at 0:50
I'll give that a shot. I find it hard to know where to start on these things. – confused Oct 10 '11 at 0:54
"pain and suffering" - my take was "algebra, sweat, and tears"... – J. M. Oct 10 '11 at 1:22