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
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1$\textbf{Hint:}$ put $x=\sin(t)$ and you are done. – user4275686 Dec 03 '14 at 10:57
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Substitute $x=\dfrac1t$ followed by $u=\sqrt{t^2-1}$. – Integreek Feb 06 '25 at 17:21
3 Answers
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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.
Listing
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$$\int \frac{dx}{\left(1-x^2\right)^{3/2}}=[x=\sin t]=\int\frac{\cos t dt}{\cos^3 t}=\int\frac{dt}{\cos^2 t}=\tan t$$
Valentin
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$= \dfrac{x}{\sqrt{1 - x^2}}$, without which the answer is incomplete. – M. Vinay Jun 18 '14 at 08:01
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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.
rajai 7
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I am referring to the first equality. Plug in $x=2$, for example... It doesn't work... – Ludolila Jun 18 '14 at 07:56
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