Thread: integration

$\displaystyle

\int x sin2x cos2x dx

$

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

$\displaystyle
\int sinucosu du

$

how would i go from here

Originally Posted by simsima_1

$\displaystyle

\int x sin2x cos2x dx

$

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

$\displaystyle
\int sinucosu du

$

how would i go from here

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

try this trig fomula

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

so you would get

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

now we have the integral

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

by parts we get...

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