# Difficult Integral

Well, you could also have started with a straight forward substitution: ##y=\arcsin x##
This means that ##x=\sin y##.
You're rid of that pesky arcsin and get easier functions to integrate.

You'd have:
$$\int (\arcsin x)^2 dx = \int y^2 d(\sin y)$$

Now do integration by parts:
$$\int y^2 d(\sin y) = y^2 \sin y - \int 2y \cos y dy$$

Repeat integration by parts, and after that back substitute x for y...

Sorry I just looked back at this thread, this is interesting, but I don't know how to follow it. Could someone explain?

Dick
Homework Helper
Sorry I just looked back at this thread, this is interesting, but I don't know how to follow it. Could someone explain?

What part don't you get? Do you have a problem with ILS's clever substitution?

What part don't you get? Do you have a problem with ILS's clever substitution?

I'm not sure on his notation with the d(siny) part particularly.

Dick
Homework Helper
I'm not sure on his notation with the d(siny) part particularly.

ILS just substituted x=sin(y). So dx becomes d(sin(y)). arcsin(x)=arcsin(sin(y))=y. dx=d(sin(y))=cos(y)dy. You might have noticed in the way I led you through there was a lot of repetition in calculating the various integrals. ILS's idea compacts that a bit.

ILS just substituted x=sin(y). So dx becomes d(sin(y)). arcsin(x)=arcsin(sin(y))=y. dx=d(sin(y))=cos(y)dy. You might have noticed in the way I led you through there was a lot of repetition in calculating the various integrals. ILS's idea compacts that a bit.

I see it, thanks. But I would probably never think of that lol.

Dick