# To integrate: $\int_{0}^{\pi}x\sin ^{2}x \cos x dx$

I have to integrate $\displaystyle \int_{0}^{\pi}x \sin^{2}x \cos x dx$ using reduction formula. I am know that for using reduction formula i have to convert my limit in $0$ to $\displaystyle \frac{\pi}{2}$, for that i using the formula that $\displaystyle \sin^{2}x= 4\sin^{2}\frac{x}{2}\cos^{2}\frac{x}{2}$ and $\displaystyle \cos x=2\cos^{2}\frac{x}{2}+1$ and then i am taking $\displaystyle \frac{x}{2}=t$ after that i am stuck. So please help me to get answer.

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I am thinking let $t=\tan(x/2)$. – Patrick Li Oct 14 '12 at 13:10
If i take $\displaystyle x=\pi$ then $t=\infty$ so how can i use reduction formula? – Kns Oct 14 '12 at 13:14

I would try to apply integration by parts, taking into account

$$\frac{d}{dx}\frac{\sin^3x}{3}=\sin^2x\cos x$$

so that

$$\int_0^\pi x\sin^2x\cos xdx=\left.x\frac{\sin^3x}{3}\right|_0^\pi-\int_0^\pi\frac{\sin^3x}{3}dx$$

The following should be easy.

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By parts: $f(x)=x$ and $g'(x)=\sin^2x\cos x$ (hence $g(x)=\frac{\sin^3x}3$). This gives $$I:=\int_0^\pi x\sin^2x\cos xdx=\left[f(x)g(x)\right]_0^{\pi}-\frac 13\int_0^\pi\sin x(1-\cos^2x)dx=\frac 13\left[\cos x\right]_0^\pi\\+\frac 13\frac{-1}3\int_0^{\pi}\frac d{dx}(\cos^3x)dx.$$

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Edit: Disregard this answer as I took the integrand to be $\,x\sin^2x\,$ instead of $\,x\sin^2x\cos x\,$ . Yet before I'll delete this post I'll leave it hanging around for a while in case somebody find it useful.

I don't know what you call "reduction formula" to in this context, but by integrating by parts we'd get:

$$u=x\,\,,\,\,u'=1$$

$$v'=\sin^2x\,\,,\,\,v=\frac{x-\sin x\cos x}{2}\;\;\;,\;\;\text{so:}$$

$$\int_0^\pi x\sin^2x\,dx=\left.\frac{x^2-x\sin x\cos x}{2}\right|_0^\pi-\frac{1}{2}\left[\int_0^\pi x\,dx+\int_0^\pi\sin x(\cos x\,dx)\right]=$$

$$=\frac{1}{2}\left[\frac{\pi^2}{2}+\left.\frac{1}{2}\sin^2x\right|_0^\pi\right]=\frac{\pi^2}{4}$$

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