I was reading a calculus book and I saw this reduction formula:
$$\int \frac{dx}{(1+x^2)^n} = \frac{1}{2n-2}\frac{x}{(x^2+1)^{n-1}}+\frac{2n-3}{2n-2}\int\frac{1}{(x^2+1)^{n-1}}dx$$
Out of curiosity I attempted to prove it, but I got stuck near the end of it.
My attempt:
Let $x=\tan(t), dx=\sec^2(t)dt$
Substituting in the original integral we get:
$$\int\frac{\sec^2(t)}{(1+\tan^2(t))^n}{dt}$$
By trig identities the integral becomes something like this:
$\int\frac{1}{[\sec^2(t)]^{n-1}}{dt}$, which is equal to $\int{\cos^{2n-2}(t)}{dt}$, then applying the reduction formula for cosine we get this thing:
$$\int{\cos^{2n-2}(t)}{dt}= \frac{1}{2n-2}·\cos^{2n-3}(t)\sin(t) + \frac{2n-3}{2n-2}·\int{{\cos}^{2n-4}(t)}{dt}$$
Then after some algebraic and trigonometric manipulations the expression looks like this:
$$\frac{1}{2n-2}·\frac{\tan(t)}{[1+\tan^{2}(t)]^{n-1}} + \frac{2n-3}{2n-2}\int{\cos}^{2n-4}(t){dt}$$
I only need to substitute $x=\tan(t)$ to get the first part of the formula, but I don't know how to manipulate $\int{{\cos}^{2n-4}(t){dt}}$ to get an expression that I can use to finish this problem.
How do I proceed, did I make a mistake, will there ever be a proof for the Riemman Hypothesis?
P.s
I tried breaking down $\int{{\cos}^{2n-4}(t){dt}}$ into $$\int{{\cos}^{2n-2}(t)\cos^{-2}(t){dt}}$$
But after playing around with that expression I get $\int\frac{1+x^{2}}{[1+x^{2}]^{n-1}}{dx}$, which doesn't match the formula
Best Answer
Use integration by parts, $$I=\int\frac{dx}{(1+x^2)^n}=\int\frac{1}{(1+x^2)^n}\cdot 1\ dx $$ $$I=\frac{1}{(1+x^2)^n}\int 1 \ dx-\int \left((-n)\frac{2x}{(1+x^2)^{n+1}}\cdot x\right)dx$$ $$I=\frac{x}{(1+x^2)^n}+2n\int \left(\frac{(1+x^2)-1}{(1+x^2)^{n+1}}x\right)dx$$ $$I=\frac{x}{(1+x^2)^n}+2n\int \left(\frac{1}{(1+x^2)^{n}}-\frac{1}{(1+x^2)^{n+1}}\right)dx$$ $$I=\frac{x}{(1+x^2)^n}+2n\int \frac{dx}{(1+x^2)^{n}}-2n\int \frac{1}{(1+x^2)^{n+1}}dx$$ $$I=\frac{x}{(1+x^2)^n}+2nI-2n\int \frac{1}{(1+x^2)^{n+1}}dx$$ $$0=\frac{x}{(1+x^2)^n}+(2n-1)I-2n\int \frac{1}{(1+x^2)^{n+1}}dx$$ $$2n\int \frac{1}{(1+x^2)^{n+1}}dx=\frac{x}{(1+x^2)^n}+(2n-1)I$$ $$\int \frac{dx}{(1+x^2)^{n+1}}=\frac{x}{2n(1+x^2)^n}+\frac{(2n-1)}{2n}\int \frac{dx}{(1+x^2)^{n}}$$ setting $n=n-1$ $$\int \frac{dx}{(1+x^2)^{n}}=\frac{x}{(2n-2)(1+x^2)^{n-1}}+\frac{(2n-3)}{2n-2}\int \frac{dx}{(1+x^2)^{n-1}}$$