- #26

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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?