# Integration by Parts practice

• Feb 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?

$\displaystyle \int x\sin x\cos xdx$

I used integration by parts with

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

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

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

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

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

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

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

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

$\displaystyle =\frac{\sin 2x-2x\cos 2x}{8}+C$
• Feb 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?

$\displaystyle \int x\sin x\cos xdx$

I used integration by parts with

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\displaystyle = -\frac{1}{2}x\cos{2x} + \frac{1}{2}\sin{2x} + C$.
• Feb 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?

$\displaystyle \int x\sin x\cos xdx$

I used integration by parts with

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

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

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

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

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

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

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

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

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

There are several ways you can check this, including:

1. Using WolframAlpha.