# Evaluating $\int \frac{1}{(1-x^2)^{3/2}} dx$

I am struggling to evaluate the following integral:
$$\int \frac{1}{(1-x^2)^{3/2}} dx$$
I tried a lot to factorize the expression but I didn't reach the solution. Please someone help me.

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Is the denominator supposed to be divided by two or is the whole term supposed to be to the $3/2$ power? It appears @azarel fixed it, but I just want to clarify. –  Joe Jun 10 '12 at 18:13

Hint:

Set $x=\sin(t)$, then everything will turn out very well.

This often helps when you have some expresion like $1-x^2$ in your integral.

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$$\int \frac{dx}{\left(1-x^2\right)^{3/2}}=[x=\sin t]=\int\frac{\cos t dt}{\cos^3 t}=\int\frac{dt}{\cos^2 t}=\tan t$$

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$= \dfrac{x}{\sqrt{1 - x^2}}$, without which the answer is incomplete. –  M. Vinay Jun 18 '14 at 8:01

Hint: Let $x=\sin\theta$, so $dx=\cos\theta\,d\theta$ and $1-x^2=cos^2\theta$.

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$\displaystyle (1-x^2)^{\frac{3}{2}} =x^3(\frac{1}{x^2}-1)^{\frac{3}{2}}$

From it you can substitute:

$\displaystyle [\frac{1}{x^2}-1] =z$

By differentiating both sides we will get $\displaystyle \frac{-2}{x^3}dx=dz$.

In this way we can also solve the integral.

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Are you sure about the equality you stated?.. –  Ludolila Jun 18 '14 at 7:49
Yes I think so. –  rajai 7 Jun 18 '14 at 7:50
Friend you can check my answer. –  rajai 7 Jun 18 '14 at 7:52
I am referring to the first equality. Plug in $x=2$, for example... It doesn't work... –  Ludolila Jun 18 '14 at 7:56
I think now you are okk with the solution.@ Ludolila. –  rajai 7 Jun 18 '14 at 10:54

$\textbf{Hint:}$ put $x=\sin(t)$ and you are done.

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