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.
Evaluating $\int \frac{1}{(1-x^2)^{3/2}} dx$
2
$\begingroup$
calculus
integration
indefinite-integrals
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0Is 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. – 2012-06-10
4 Answers
6
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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0$= \dfrac{x}{\sqrt{1 - x^2}}$, without which the answer is incomplete. – 2014-06-18
0
$\displaystyle (1-x^2)^{\frac{3}{2}} =x^3(\frac{1}{x^2}-1)^{\frac{3}{2}}$
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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0I think now you are okk with the solution.@ Ludolila. – 2014-06-18
0
$\textbf{Hint:}$ put $x=\sin(t)$ and you are done.