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I have to evaluate $$\int\frac{1}{(1-u^2)^\frac{3}{2}}du$$ I tried to attack this integral changing variable $$1-u^2 =\sin^2\phi$$ but in this way i reconduct myself to evaluate $$\int \frac{1}{\sin^3\phi}\sin\phi d\phi$$

Integreek
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Rosario Di Mari
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    Consider $\int\frac{\sin x}{\sin^4 x}dx$, then ... – Bowei Tang Oct 23 '24 at 21:43
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    After the edit, consider $\int \csc^2 xdx=\int \sec^2(\pi/2-x)dx$ and $(\tan x)'=\sec^2 x$ – Bowei Tang Oct 23 '24 at 21:47
  • Do I have to change variable furthermore? – Rosario Di Mari Oct 23 '24 at 21:54
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    Sorry, I don't understand this comment. Do you mean: 1) "was it necessary to let $1-u^2=\sin^2\phi$?", or 2) "will I have to do another change of variable after that?"? If it is 2, please look at the linked post. If it is 1, this other post is relevant: https://math.stackexchange.com/questions/156585 – Anne Bauval Oct 23 '24 at 22:00

3 Answers3

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Another way is by perform hyperbolic substitution.

Under $u = \tanh\left(v\right) \to \mathrm{d}u = \operatorname{sech}^{2}\left(v\right) \, \mathrm{d}v $ , we get :

$$I= \int \frac{\operatorname{sech}^{2}\left(v\right)}{\left(1 - \tanh^{2}\left(v\right)\right)^{\frac{3}{2}}} \, \mathrm{d}v= \int \frac{1}{\operatorname{sech}\left(v\right)} \, \mathrm{d}v= \sinh\left(v\right)$$

And in terms of $u$ :

$$I=\sinh\left(\operatorname{artanh}\left(u\right)\right) = \frac{u}{\sqrt{1 - u} \sqrt{u + 1}}+C=\frac{u}{\sqrt{1 - u^{2}}} + C$$

Antony Theo.
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observe that $(\cot\phi)' = -\csc^2\phi \\$

$$\therefore \int -\csc^2\phi \,d\phi = \cot\phi + C = \frac{u}{\sqrt{1-u^2}} + C$$

But may I suggest an easier solution:

$$I=\int\frac{1}{(1-u^2)^\frac{3}{2}}\,du = \int\frac{1}{u^3({\frac{1}{u^2}-1})^\frac{3}{2}}\,du = \frac{-1}{2}\int\frac{-2}{u^3({\frac{1}{u^2}-1})^\frac{3}{2}}\,du $$

substituting ${\frac{1}{u^2}-1} = t $ gives

$$I=\frac{-1}{2}\int\frac{1}{t^\frac{3}{2}}\,dt = t^\frac{-1}{2} +C = \frac{u}{\sqrt{1-u^2}} + C$$

Edmaand Vlad Phiyik
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    Thank you very much:) very useful. I need it to calculate the speed of a body uniformely accelerated in special relativity ;) – Rosario Di Mari Oct 25 '24 at 10:39
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Here's a generalized approach $$I=\int x^m(a+bx^n)^p\,dx$$

Case (1):

If $p$ is an integer, substitute $x=t^N$, where $N$ is LCM of $m,n$.

Case (2):

If $\frac{m+1}{n}$ is an integer, substitute $a+bx^n=t^N$, where $N$ is denominator of $p$.

Case (3):

If $\frac{m+1}{n}+p$ is an integer, substitute $ax^{-1}+b=t^N$, where $N$ is denominator of $p$.

Given integral;

$$I=\int (1-u^2)^{-\frac{3}{2}}du$$

Your integral matches case (3); hence substitute $\color{red}{u^{-1}-1=t^2}$

After some simplification work, you end up on,

$$I=-\int \frac{(2t)(t^2+1)}{((t^2+1)^2-1)^{\frac{3}{2}}}\,dt$$

Do another substitution of $t^2+1=z$, then $z^2-1=x$ and simplify your way back (using standard well known integrals) and you will end up on,

$$I=\frac{u}{\sqrt{1-u^2}}+C$$

Amrut Ayan
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