# Calculus integration problem: $\int \sin^5 (x) \cos^2 (x)\,dx$

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
Similar integral: math.stackexchange.com/questions/902722/… – Martin Sleziak Aug 19 '14 at 11:01

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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$$\int \sin^5 (x) \cos^2(x) dx$$ $$= \int(\sin^2(x))^2 \cos^2(x) \sin(x) dx$$ $$=-\int(1 - \cos^2(x))^2 cos^2(x) (-sin(x) dx)$$

Let $u = \cos(x)$ $\implies du = -\sin(x) dx$

$$= -\int(1 - u^2)² u² (du)$$ $$= -\int(1 - 2u^2 + u^4) u^2 du$$ $$= -\int(u^2 - 2u^4+ u^6) du$$ $$= -\left(\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7}\right) + C$$ $$= -u^3\left(\frac{1}{3} - \frac{2u^2}{5} +\frac{ u^4}{7}\right) + C$$ $$= -\cos^3(x) \left(\frac{1}{3} - \frac{2\cos^2(x)}{5} + \frac{\cos^4(x)}{7}\right) + C$$ $$= -\cos^3(x)\frac{15\cos^4(x) - 42\cos^2(x) + 35}{105} + C$$

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For some basic information about writing math at this site see e.g. here, here, here and here. – SBareS Oct 21 '15 at 14:46

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