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I'm having difficulty proving this integral reduction formula by parts

If

$$I_n=\int\frac{dx}{(a^2-x^2)^n}$$

, then

$$\int\frac{dx}{(a^2-x^2)^n}=\frac{x}{2a^2(n-1)(a^2-x^2)^{n-1}}+\left(\frac{2n-3}{2a^2(n-1)}\right)I_{n-1}$$

I managed to do it using a trigonometric substitution:

$$\int\frac{dx}{(a^2-x^2)^n}\space\begin{vmatrix}x=a\sin(\theta)\\dx=a\cos(\theta)d\theta\end{vmatrix}=\int\frac{a\cos(\theta)}{(a^2- a^2\sin^2(\theta))^n}d\theta=\int\frac{a\cos(\theta)}{a^{2n}\cos^{2n}(\theta)}d\theta\\=\frac{1}{a^{2n-1}}\int\sec^{2n-1}(\theta)d\theta$$

Using the reduction formula

$$\int\sec^n(\theta)d\theta=\frac{1}{n-1}\sec^{n-2}(\theta)\tan(\theta)+\left(\frac{n-2}{n-1}\right)\int\sec^{n-2}(\theta)d\theta$$

this integral becomes

$$\frac{1}{a^{2n-1}}\left[\frac{1}{2(n-1)}\sec^{2n-3}(\theta)\tan(\theta)+\left(\frac{2n-3}{2(n-1)}\right)\int\sec^{2n-3}(\theta)d\theta\right]$$

Based on the substitution $x=a\sin(\theta)$ and $dx=a\cos(\theta)d\theta$:

$$\sec(\theta)=\frac{a}{\sqrt{a^2-x^2}}\quad\tan(\theta)=\frac{x}{\sqrt{a^2-x^2}}\quad d\theta=\frac{dx}{\sqrt{a^2-x^2}}$$

So

$$\int\frac{dx}{(a^2-x^2)^n}=\frac{1}{a^{2n-1}}\left[\frac{1}{2(n-1)}\sec^{2n-3}(\theta)\tan(\theta)+\left(\frac{2n-3}{2(n-1)}\right)\int\sec^{2n-3}(\theta)d\theta\right]\\=\frac{1}{a^{2n-1}}\bigg[\frac{1}{2(n-1)}\bigg(\frac{a}{\sqrt{a^2-x^2}}\bigg)^{2n-3}\bigg(\frac{x}{\sqrt{a^2-x^2}}\bigg)+\left(\frac{2n-3}{2(n-1)}\right)\int\bigg(\frac{a}{\sqrt{a^2-x^2}}\bigg)^{2n-3}\bigg(\frac{dx}{\sqrt{a^2-x^2}}\bigg)\bigg]\\=\frac{1}{a^{2n-1}}\bigg[\frac{a^{2n-3}x}{2(n-1)(\sqrt{a^2-x^2})^{2n-2}}+\left(\frac{2n-3}{2(n-1)}\right)\int\frac{a^{2n-3}}{(\sqrt{a^2-x^2})^{2n-2}}dx\bigg]\\=\frac{1}{a^{2n-1}}\left[\frac{a^{2n-3}x}{2(n-1)(a^2-x^2)^{n-1}}+\left(\frac{2n-3}{2(n-1)}\right)\int\frac{a^{2n-3}}{(a^2-x^2)^{n-1}}dx\right]\\=\frac{x}{2a^2(n-1)(a^2-x^2)^{n-1}}+\left(\frac{2n-3}{2a^2(n-1)}\right)\int\frac{dx}{(a^2-x^2)^{n-1}}\\=\frac{x}{2a^2(n-1)(a^2-x^2)^{n-1}}+\left(\frac{2n-3}{2a^2(n-1)}\right)I_{n-1}$$

However I also wanted to prove this using integration by parts but I seem to be stuck:

$$\int\frac{dx}{(a^2-x^2)^n}=\int\frac{x}{x(a^2-x^2)^n}dx\space\begin{vmatrix}u=\frac{1}{x}\\du=-\frac{1}{x^2}dx\end{vmatrix}dv=\frac{x}{(a^2-x^2)^n}dx\\v=\int\frac{x}{(a^2-x^2)^n}dx\begin{vmatrix}u=a^2-x^2\\du=-2xdx\end{vmatrix}v=-\frac{1}{2}\int\frac{du}{u^n}=-\frac{1}{2}\left(\frac{u^{-n+1}}{-n+1}\right)\\=-\frac{1}{2(1-n)u^{n-1}}=-\frac{1}{2(1-n)(a^2-x^2)^{n-1}}\\\int udv=uv-\int vdu\\\int\frac{dx}{(a^2-x^2)^n}=-\frac{1}{2x(1-n)(a^2-x^2)^{n-1}}-\frac{1}{2(n-1)}\int\frac{1}{x^2(a^2-x^2)^{n-1}}dx$$

Don't know how to proceed from here, any help?

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Differentiation helps us to reduce calculation

$$\dfrac{d\{x(a^2-x^2)^n\}}{dx}a$$

$$=(a^2-x^2)^n-2nx^2(a^2-x^2)^{n-1}=(a^2-x^2)^n+2n(a^2-x^2-a^2)(a^2-x^2)^{n-1}$$

$$\implies\dfrac{d\{x(a^2-x^2)^n\}}{dx}=(1+2n)(a^2-x^2)^n-2na^2(a^2-x^2)^{n-1}$$

Integrating both sides, $$x(a^2-x^2)^n=(2n+1)I_n-2na^2I_{n-1}$$

Set $n=-m$

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$$I_n=\int\frac{dx}{(ax^2+b)^n}$$ $$dv=dx\Rightarrow v=x\\ u=\frac1{(ax^2+b)^n}\Rightarrow du=\frac{-2anx}{(ax^2+b)^{n+1}}dx$$ $$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2}{(ax^2+b)^{n+1}}dx$$ $$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bn\int\frac{dx}{(ax^2+b)^{n+1}}$$ $$I_n=\frac{x}{(ax^2+b)^n}+2n\int\frac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$ $$I_n=\frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$ $$2bnI_{n+1}=\frac{x}{(ax^2+b)^n}+(2n-1)I_n$$ $$I_{n+1}=\frac{x}{2bn(ax^2+b)^n}+\frac{2n-1}{2bn}I_n$$ replacing $n+1$ with $n$, $$I_{n}=\frac{x}{2b(n-1)(ax^2+b)^{n-1}}+\frac{2n-3}{2b(n-1)}I_{n-1}$$ For your integral, set $a=-1$ and replace $b$ with $b^2$

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  • $\begingroup$ But is it possible to do it through an independent IBP process, an not from a different reduction formula? $\endgroup$
    – Anson Pang
    Jan 3 '19 at 3:15
  • $\begingroup$ @AnsonPang yes, but the process would be exactly the same. $\endgroup$
    – clathratus
    Jan 3 '19 at 3:18
  • $\begingroup$ Oh, I see, thanks. $\endgroup$
    – Anson Pang
    Jan 3 '19 at 3:19

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