Vtgyjfy

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

If

$$I_n=intfrac{dx}{(a^2-x^2)^n}$$

, then

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

I managed to do it using a trigonometric substitution:

$$intfrac{dx}{(a^2-x^2)^n}spacebegin{vmatrix}x=asin(theta)\dx=acos(theta)dthetaend{vmatrix}=intfrac{acos(theta)}{(a^2 a^2sin^2(theta))^n}dtheta=intfrac{acos(theta)}{a^{2n}cos^{2n}(theta)}dtheta\=frac{1}{a^{2n-1}}intsec^{2n-1}(theta)dtheta$$

Using the reduction formula

$$intsec^n(theta)dtheta=frac{1}{n-1}sec^{n-2}(theta)tan(theta)+left(frac{n-2}{n-1}right)intsec^{n-2}(theta)dtheta$$

this integral becomes

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

Based on the substitution $x=asin(theta)$ and $dx=acos(theta)dtheta$:

$$sec(theta)=frac{a}{sqrt{a^2-x^2}}quadtan(theta)=frac{x}{sqrt{a^2-x^2}}quad dtheta=frac{dx}{sqrt{a^2-x^2}}$$

So

$$intfrac{dx}{(a^2-x^2)^n}=frac{1}{a^{2n-1}}left[frac{1}{2(n-1)}sec^{2n-3}(theta)tan(theta)+left(frac{2n-3}{2(n-1)}right)intsec^{2n-3}(theta)dthetaright]\=frac{1}{a^{2n-1}}bigg[frac{1}{2(n-1)}bigg(frac{a}{sqrt{a^2-x^2}}bigg)^{2n-3}bigg(frac{x}{sqrt{a^2-x^2}}bigg)+left(frac{2n-3}{2(n-1)}right)intbigg(frac{a}{sqrt{a^2-x^2}}bigg)^{2n-3}bigg(frac{dx}{sqrt{a^2-x^2}}bigg)bigg]\=frac{1}{a^{2n-1}}bigg[frac{a^{2n-3}x}{2(n-1)(sqrt{a^2-x^2})^{2n-2}}+left(frac{2n-3}{2(n-1)}right)intfrac{a^{2n-3}}{(sqrt{a^2-x^2})^{2n-2}}dxbigg]\=frac{1}{a^{2n-1}}left[frac{a^{2n-3}x}{2(n-1)(a^2-x^2)^{n-1}}+left(frac{2n-3}{2(n-1)}right)intfrac{a^{2n-3}}{(a^2-x^2)^{n-1}}dxright]\=frac{x}{2a^2(n-1)(a^2-x^2)^{n-1}}+left(frac{2n-3}{2a^2(n-1)}right)intfrac{dx}{(a^2-x^2)^{n-1}}\=frac{x}{2a^2(n-1)(a^2-x^2)^{n-1}}+left(frac{2n-3}{2a^2(n-1)}right)I_{n-1}$$

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

$$intfrac{dx}{(a^2-x^2)^n}=intfrac{x}{x(a^2-x^2)^n}dxspacebegin{vmatrix}u=frac{1}{x}\du=-frac{1}{x^2}dxend{vmatrix}dv=frac{x}{(a^2-x^2)^n}dx\v=intfrac{x}{(a^2-x^2)^n}dxbegin{vmatrix}u=a^2-x^2\du=-2xdxend{vmatrix}v=-frac{1}{2}intfrac{du}{u^n}=-frac{1}{2}left(frac{u^{-n+1}}{-n+1}right)\=-frac{1}{2(1-n)u^{n-1}}=-frac{1}{2(1-n)(a^2-x^2)^{n-1}}\int udv=uv-int vdu\intfrac{dx}{(a^2-x^2)^n}=-frac{1}{2x(1-n)(a^2-x^2)^{n-1}}-frac{1}{2(n-1)}intfrac{1}{x^2(a^2-x^2)^{n-1}}dx$$

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

$$I_n=intfrac{dx}{(ax^2+b)^n}$$
$$dv=dxRightarrow v=x\ u=frac1{(ax^2+b)^n}Rightarrow du=frac{-2anx}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2}{(ax^2+b)^{n+1}}dx$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{ax^2+b}{(ax^2+b)^{n+1}}dx-2bnintfrac{dx}{(ax^2+b)^{n+1}}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nintfrac{dx}{(ax^2+b)^{n}}-2bnI_{n+1}$$
$$I_n=frac{x}{(ax^2+b)^n}+2nI_n-2bnI_{n+1}$$
$$2bnI_{n+1}=frac{x}{(ax^2+b)^n}+(2n-1)I_n$$
$$I_{n+1}=frac{x}{2bn(ax^2+b)^n}+frac{2n-1}{2bn}I_n$$
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 $b^2$

Differentiation helps us to reduce calculation

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

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

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

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

Set $n=-m$