1. ## Integrating by parts

I have a problem that I've been working on:

Consider the definite integral

I have:

arcsin(4x)(x^2/2) minus the integral [IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG]

I'm really not sure where to go from here.

2. Originally Posted by C.C.
I have a problem that I've been working on:

Consider the definite integral

I have:

arcsin(4x)(x^2/2) minus the integral [IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-1.jpg[/IMG][IMG]file:///C:/Users/Tawny/AppData/Local/Temp/moz-screenshot-2.jpg[/IMG]

I'm really not sure where to go from here.
If $\displaystyle x=\frac{1}{4}\sin(u)$, then $\displaystyle dx=\frac{1}{4}\cos(u)$

Since $\displaystyle u=\arcsin(4x)$, the new bounds are $\displaystyle 0...\frac{\pi}{2}$

So we have $\displaystyle \frac{1}{16}\int_0^{\pi/2} u\sin u\cos u\,du=\frac{1}{32}\int_0^{\pi/2}u\sin2u\,du$

From here it's an easy integration by parts. I calculate the final answer to be $\displaystyle \frac{\pi}{128}$.