# Annoyingly difficult integral

1. Feb 21, 2007

### mwaso

The problem statement, all variables and given/known data

integrate dx/(1-x^2)^(3/2)

The attempt at a solution

I first noticed that the derivative of inverse sin equals (1 x^2)^(-1/2) and that my equation is that derivative cubed. However, I had no idea what to do with the cube, so I dicarded that thought.

Next I tried to do a u substitution. obviously I can't make u equal 1-x^2 because that leaves me with an extra x in my du that won't go away. I was inspired by remembering some powers of trig integrations to try to make du equal the derivative of inverse sin, but that also failed as I had no inverse sin in the original equation to replace with u and last I checked, I can't integrate du^3.

I finally checked my tables of integrals and found where the integral of du/(a^2-u^2)^(3/2) equals u/a^2(a^2-u^2). This seems to give me my answer, but I can't figure out how to derive that formula.

2. Feb 21, 2007

### benorin

From $$I=\int\frac{dx}{(1-x^2)^{\frac{3}{2}}}$$ apply the substitution $$x=\sin\theta$$. Hint: you should get $$I=\int\sec ^2 \theta\, d\theta$$.

3. Feb 21, 2007

### theperthvan

You probably substituted wrongly to get du^3