# Integration by Parts practice

• February 20th 2010, 09:11 PM
Keithfert488
Integration by Parts practice
Hi I was wondering if I got this right so could someone check for me?

$\int x\sin x\cos xdx$

I used integration by parts with

$u=\sin x\cos x$ and $dv=xdx$

and got... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}\int x^2\cos 2xdx$

Then, I used parts again with $u=x^2$ and $dv=\cos 2xdx$

... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}-\int x\sin 2xdx)$

Parts yet again with $u=x$ and $dv=\sin 2xdx$

... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}+(\frac{x\cos 2x}{2}-\frac{1}{2}\int \cos 2xdx))$

$=\frac{x^2\sin 2x}{4}-\frac{1}{2}(\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4})+C$

$=\frac{x^2\sin 2x}{4}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C$

$=\frac{\sin 2x-2x\cos 2x}{8}+C$
• February 20th 2010, 09:36 PM
Prove It
Quote:

Originally Posted by Keithfert488
Hi I was wondering if I got this right so could someone check for me?

$\int x\sin x\cos xdx$

I used integration by parts with

$u=\sin x\cos x$ and $dv=xdx$

and got... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}\int x^2\cos 2xdx$

Then, I used parts again with $u=x^2$ and $dv=\cos 2xdx$

... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}-\int x\sin 2xdx)$

Parts yet again with $u=x$ and $dv=\sin 2xdx$

... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}+(\frac{x\cos 2x}{2}-\frac{1}{2}\int \cos 2xdx))$

$=\frac{x^2\sin 2x}{4}-\frac{1}{2}(\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4})+C$

$=\frac{x^2\sin 2x}{4}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C$

$=\frac{\sin 2x-2x\cos 2x}{8}+C$

Rewrite the integral as $\frac{1}{2}\int{x\sin{2x}\,dx}$.

Let $u = x$ so that $du = 1$

Let $dv = \sin{2x}$ so that $v = -\frac{1}{2}\cos{2x}$.

So $\frac{1}{2}\int{x\sin{2x}\,dx} = -\frac{1}{2}x\cos{2x} - \int{-\frac{1}{2}\cos{2x}\cdot 1\,dx}$

$= -\frac{1}{2}x\cos{2x} + \frac{1}{2}\int{\cos{2x}\,dx}$

$= -\frac{1}{2}x\cos{2x} + \frac{1}{2}\sin{2x} + C$.
• February 20th 2010, 09:36 PM
mr fantastic
Quote:

Originally Posted by Keithfert488
Hi I was wondering if I got this right so could someone check for me?

$\int x\sin x\cos xdx$

I used integration by parts with

$u=\sin x\cos x$ and $dv=xdx$

and got... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}\int x^2\cos 2xdx$

Then, I used parts again with $u=x^2$ and $dv=\cos 2xdx$

... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}-\int x\sin 2xdx)$

Parts yet again with $u=x$ and $dv=\sin 2xdx$

... $\frac{x^2\sin x\cos x}{2}-\frac{1}{2}(\frac{x^2\sin x}{2}+(\frac{x\cos 2x}{2}-\frac{1}{2}\int \cos 2xdx))$

$=\frac{x^2\sin 2x}{4}-\frac{1}{2}(\frac{x^2\sin 2x}{2}+\frac{x\cos 2x}{2}-\frac{\sin 2x}{4})+C$

$=\frac{x^2\sin 2x}{4}-\frac{x^2\sin 2x}{4}-\frac{x\cos 2x}{4}+\frac{\sin 2x}{8}+C$

$=\frac{\sin 2x-2x\cos 2x}{8}+C$

There are several ways you can check this, including:

1. Using WolframAlpha.