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Using only simple calculus and rewriting the integral how could I find the recursive integral of

$I\left(n\right)=\int \frac{1}{\left(1-x^2\right)^n}\:dx$

I tried using trig substitution like so:

$x = \sin(u)$ with $dx = \cos(u) du$

writing $1 = \sin^2(u) + \cos^2(u)$

and also $(1-\sin(u)^2)=\cos^2(u)$

Eventually I reach that $I(n) = \int \frac{\sin^2\left(u\right)}{\cos^{n-1}du}du + \int \cos^{1-n}\left(u\right)du$

I stopped here as I do not know how to continue or if the route I chose is even ok.

ViktorStein
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Alex Mihoc
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2 Answers2

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For $n \ge 2$ we have $$\begin{align}\int\frac{1}{(1-x^2)^n}dx&=\int\frac{1-x^2+x^2}{(1-x^2)^n}dx\\ &=\int\frac{1}{(1-x^2)^{n-1}}dx+\int x\cdot\frac{x}{(1-x^2)^n}dx\end{align}.$$ For the second integral, use integration by parts with $u=x, dv=\frac{x}{(1-x^2)^n}dx$: $$\int x\cdot\frac{x}{(1-x^2)^n}dx=x\cdot\frac{1}{2(n-1)(1-x^2)^{n-1}}-\frac{1}{2(n-1)}\int\frac{1}{(1-x^2)^{n-1}}dx $$ so we get $$\begin{align}I(n)&=I(n-1)+\frac{x}{2(n-1)(1-x^2)^{n-1}}-\frac{1}{2(n-1)}I(n-1)\\ &=\frac{2n-3}{2(n-1)}I(n-1)+\frac{x}{2(n-1)(1-x^2)^{n-1}}\end{align} $$

ViktorStein
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bjorn93
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Integrate by parts

\begin{align}\int\frac{1}{(1-x^2)^n}dx= I_n =&\int\frac1{2(n-1)x^{2n-3}}\ d\bigg( \frac{x^{2(n-1)}}{(1-x^2)^{n-1}}\bigg)\\ = &\ \frac{x}{2(n-1)(1-x^2)^{n-1}}+ \frac{2n-3}{2(n-1)}I_{n-1} \end{align}

Quanto
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