Evaluate $\int_{0}^{∞}(1+x^2)^{−(m+1)}dx$ where m is a natural number. 
QUESTION: Evaluate $$\int_{0}^{∞}(1+x^2)^{−(m+1)}dx$$ where m is a natural number.


MY ANSWER: Observing the question, the first substitution I made is changing $x$ to $tan\theta$, since $1+tan^2\theta=sec^2\theta$, this should lead me somewhere.
Now, changing the limits of the integration and after some easy calculation, I arrive at-
$$\int_{0}^{\frac{π}2}(cos\theta)^{2m}d\theta$$ Since, we do not know that value of $m$ how do we solve such an integration?
I am stuck here. Any answers or alternative methods are much appreciated. Thank you so much.
 A: In terms of the Beta function,$$\int_0^{\pi/2}\cos^{2m}\theta d\theta=\frac12\operatorname{B}\left(m+\frac12,\,\frac12\right)=\frac{\Gamma\left(m+\frac12\right)\Gamma\left(\frac12\right)}{2\Gamma(m+1)}=\frac{(2m)!}{m!^22^{2m+1}}\pi.$$
A: Using the fact that $\cos (x) = (e^{ix} + e^{-ix})/2$, expand your integrand to obtain
$$\frac{2^{-2m}}{4}\sum_{k=0}^{2m} {2m \choose k}\int_0^{2\pi} e^{i2kx} e^{-i2mx}\,dx.$$
But the orthogonality of the complex exponentials on the interval $[0, 2\pi]$ immediately tells us that the integral is proportional to the kronecker delta as $2\pi\delta_{m, k}$. This your integral reduces to 
$$\pi\,2^{-2m+1}{2m \choose m}.$$
A: Denote
$$I_m= \int_{0}^{\infty}\frac{1}{(1+x^2)^{m+1}}\ dx$$
You have $I_0= \frac{\pi}{2}$ and with integration by parts
$$\begin{aligned}I_m &= \int_{0}^{\infty}\frac{1+x^2}{(1+x^2)^{m+2}}\ dx = I_{m+1} + \int_{0}^{\infty}\frac{x^2}{(1+x^2)^{m+2}}\ dx\\
&=I_{m+1}+\left[x \left(\frac{-1}{2m+2}\frac{1}{(1+x^2)^{m+1}}\right)\right]_0^\infty + \frac{1}{2m+2} I_m
\end{aligned}$$
Therefore the relation
$$I_{m+1} =  \frac{2m+1}{2m+2} I_m$$
From there you can compute $I_m$.
