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What's the integration of $$\int \sin^5 (x) \cos^2 (x)\,dx?$$

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Share with us what you've tried so we can help you along. –  JohnD Feb 3 '13 at 19:53
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Hint: Write $$ \sin^5(x)\cos^2(x)=(\sin^2(x))^2\cos^2(x)\sin(x). $$

Now use $\cos^2(x)+\sin^2(x)=1$ and do the appropriate change of variable.

This is the general method to integrate functions of the type $$ \cos^n(x)\sin^m(x) $$ when one of the integers $n,m$ is odd.

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Using trig identities, you can show that: $$\sin ^5(x) \cos ^2(x)=\frac{5 \sin (x)}{64}+\frac{1}{64} \sin (3 x)-\frac{3}{64} \sin (5 x)+\frac{1}{64} \sin (7 x)$$

To do this, first use the "Power-reduction formulas" to reduce to get: $$\sin^5(x)=\frac{10 \sin x - 5 \sin 3 x+ \sin 5 x}{16}$$ $$\cos^2(x)=\frac{1 + \cos (2 x)}{2}$$ And then use: $$\cos (2 x) \sin (nx) = {{\sin((n+2)x) - \sin((n-2)x)} \over 2}$$

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Could you show how? As I see it now it may as well just be reverse-engineered from the Wolfram Alpha solution. I'm not saying you did this, but I don't know what identities you used to come to this. –  user50407 Feb 3 '13 at 20:07
    
Is it really how you solve the question in general, or is it just to provide a different answer? Just curious. –  1015 Feb 3 '13 at 20:21
    
@MichaelCorleone - fixed. –  nbubis Feb 3 '13 at 20:21
    
Yes I saw it and upvoted it. Thanks –  user50407 Feb 3 '13 at 20:22
    
@julien - just to provide a different answer of course :) It's not a very practical way, but it's helpful if the powers are very large, so that you can use general power reduction rules. –  nbubis Feb 3 '13 at 20:22
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