Integrating by parts

**C.C.** · November 9th 2009, 08:49 PM

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.

**redsoxfan325** · November 9th 2009, 09:27 PM

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 $x=\frac{1}{4}\sin(u)$ , then $dx=\frac{1}{4}\cos(u)$

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

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

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

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