Integration by Substitution

**kendon** · August 3rd 2006, 01:12 AM

Trying to integrate (arcsin(x))^2/(sqrt 1-x^2)
So far i have made
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.
Thanks in advance
Ken

**CaptainBlack** · August 3rd 2006, 03:00 AM

Originally Posted by kendon

Trying to integrate (arcsin(x))^2/(sqrt 1-x^2)
So far i have made
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.
Thanks in advance
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:

$<br /> \int u^2\ du<br />$ .

Look OK to me.

Now:

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

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

RonL

**galactus** · August 3rd 2006, 03:01 AM

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?.

**kendon** · August 3rd 2006, 04:04 AM

thanks a lot that has been superb.
cheers
ken

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