integration

**simsima_1** · Mar 13th 2008, 01:26 PM

$ \int x sin2x cos2x dx $

i started of by making u=2x, du=x, which will cancel out the x and give me

$ \int sinucosu du $

how would i go from here

**TheEmptySet** · Mar 13th 2008, 01:40 PM

Originally Posted by simsima_1

$ \int x sin2x cos2x dx $

i started of by making u=2x, du=x, which will cancel out the x and give me

$ \int sinucosu du $

how would i go from here

if u=2x then du=2dx not x sorry

try this trig fomula

$2\sin(\theta)\cos(\theta)=\sin(2\theta)$

so you would get

$\sin(2x)\cos(2x)=\frac{1}{2}\sin(4x)$

now we have the integral

$\frac{1}{2}\int x \sin(4x)dx$

by parts we get...

$\frac{1}{2}\int x \sin(4x)dx=\frac{1}{2}[-\frac{x}{4}\cos(4x)+\frac{1}{16}\sin(4x)]=-\frac{x}{8}\cos(4x)+\frac{1}{32}\sin(4x)+C$

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