Integration of $x^2 \sin(x)$ by parts

Question

How would I integrate the following?

$$\int_0^{\pi/2} x^2\sin(x)\,dx$$

I did $u=x^2$ and $dv=\sin(x)$

I got

$x^2-\cos(x)+2\int x\cos(x)\,dx.\quad$ I then used $u=x$ and $dv=\cos(x).$

I got

$$x^2-\cos(x)+2[x-\sin(x)-\int\sin(x)]$$

then

$x^2-\cos(x)+-2 \sin(x)(x)-\cos(x)\Big|_0^{\pi/2} =\dfrac{\pi^2}{4}-0-2$

@Fernando Martinez consider changing the titles of your last 3 question to something more concrete, such as How to integrate $x^2\sin(x)$ by part. This helps future users who could have the same question — Jean-Sébastien, Sep 06 '13 at 21:36

amWhy · Accepted Answer · 2013-09-06T21:30:00.430

3

You need to multiply $u$ and $v$, then subtract the subsequent integral:

So you should have $$\begin{align} \int_0^{\pi/2} x^2\sin(x)\,dx & = -x^2\cos(x)+2\int x\cos(x)\,dx \\ \\ & = -x^2 \cos x + 2\Big[x \sin x - \int \sin x\,dx\Big]\\ \\ & = -x^2\cos x + 2x \sin x - (-2\cos x)\\ \\ & = -x^2 \cos x + 2x \sin x + 2\cos x \Big|_0^{2\pi}\end{align}$$

And proceed from there.

edited Sep 06 '13 at 21:30

answered Sep 06 '13 at 21:08

amWhy

209,954

But would not the negative two turn positive because dv=sin(x) and then $v=-\cos(x)$ – Fernando Martinez Sep 06 '13 at 21:10
Yes, I had a typo...It looked like you were subtracting $\cos x$ from $x^2$, etc. – amWhy Sep 06 '13 at 21:11
@amWhy: You have been on an integration tear today! +1 – Amzoti Sep 07 '13 at 01:54

score 3 · Answer 2 · edited Feb 25 '15 at 16:08

3

$u=x^2$ then $du=2xdx$ and $dv=\sin(x)dx$ then $v=-\cos(x)$. So integral becomes $=-(x^2\cos(x))+2\int x\cos(x)dx$ integration by parts for $$\int x\cos(x)dx$$ $x=u, dx=du$ and $dv=\cos(x)dx, v=\sin(x)$ then $$=-(x^2\cos(x))-2\int \sin(x)dx+2x\sin(x)$$ the integral of $\sin x$ is $-\cos x$ so we get $$=2\cos(x)-x^2\cos(x)+2x\sin(x)$$

edited Feb 25 '15 at 16:08

dustin

8,241

answered Sep 06 '13 at 21:16

Ömer

2,038

Integration of $x^2 \sin(x)$ by parts

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