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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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    @BillCook: That uses the reduction formula though. I'd rather stick to identities and such.2011-10-10

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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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    There are multiple ways of writing a given function sometimes, so most likely they are the same function but in different forms.2011-10-10
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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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    "pain and suffering" - my take was "algebra, sweat, and tears"...2011-10-10