1. Integration by Substitution

Trying to integrate (arcsin(x))^2/(sqrt 1-x^2)
arcsin x = u and
du=1/(sqrt1-x^2).dx

I am at u^2 * du

I have spent hours on this and it appears obvious but i just don't get it! sorry
Any pointers i would be extremely grateful. I feel its something to do with the calculus relationship between the values of arcsin and du.
Ken

2. Originally Posted by kendon
Trying to integrate (arcsin(x))^2/(sqrt 1-x^2)
arcsin x = u and
du=1/(sqrt1-x^2).dx

I am at u^2 * du

I have spent hours on this and it appears obvious but i just don't get it! sorry
Any pointers i would be extremely grateful. I feel its something to do with the calculus relationship between the values of arcsin and du.
Ken
$\int \frac{(\arcsin(x))^2}{\sqrt{1-x^2}} dx$

you make the substitution $\arcsin(x)=u$, find $du=\frac{1}{\sqrt{1-x^2}}$, so the integral becomes:

$
\int u^2\ du
$
.

Look OK to me.

Now:

$
\int u^2\ du=\frac{u^3}{3}+C=\frac{(\arcsin(x))^3}{3}+C
$

which can be checked by differentiating it, you should get the integrand back.

RonL

3. You've got the right substitution.

Let $u=sin^{-1}(x)$

$du=\frac{1}{\sqrt{1-x^{2}}}dx$

You have:

$\int{u^{2}}du$

Now, it's easy, ain't it?.

4. thanks a lot that has been superb.
cheers
ken