Rewrite integral $S = \int^\infty_1 \frac{1}{1+x^2}dx$ using $x = tan(t)$ The following integral $S = \int^\infty_1 \frac{1}{1+x^2}dx$ needs to be transformed using $x = tan(t)$ into a new expression which has a new interval and which can be solved analytically.
The new interval I got it and it is from x = 0 to x = 1 using the improper integral identity. But filling in $x = tan(t)$ gives $S_2 = \int^1_0 \frac{1}{1+tan^2(x)}dx$.
Is the above correct and if yes how can I solve it? Integration by parts?
 A: Note $\tan \frac \pi 4 = 1$ and $\tan t \to \infty$ as $t \to \frac{\pi}{2}^-$ so the limits on your integral should be $\frac \pi 4$ and $\frac \pi 2$. You also left out the differential $\sec^2 t$ when you computed $dt$, so you should have $$\int_1^\infty \frac{1}{1+x^2} \, dx = \int_{\pi/4}^{\pi/2} \frac{1}{1 + \tan^2 t} \sec^2 t \, dt.$$
This is pretty simple once you recognize $1 + \tan^2 t = \sec^2 t$.
A: no, that is not correct, but you are on the right track. first off one needs to express the improper integral as a limit $\lim_{b\to\infty}\int_1^b\frac{dx}{1+x^2}$. Now $x(t)=\tan t$ is a function of $t$, not $x$ as you have it in the integrand. What is $dx$ in terms of $t$? If $x=1$, what is $t$( for the new lower bound)? If $x=b$, what is $t$ for the new upper bound?
after you make the appropriate substitutions, think about how to integrate the resulting integral that is expressed in terms of $t$
A: we have $$1+\tan(t)^2=\frac{\cos(t)^2+\sin(t)^2}{\cos(t)^2}=\frac{1}{\cos(t)^2}$$
and $$dx=\frac{1}{\cos(t)^2}dt$$
A: If $x = \tan t$, $dx = \sec^2 t \ dt = (1+\tan^2 t) \ dt$. Thus the integral is 
$$\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} dt = \frac{\pi}{4}$$
Honestly, though, you should just have memorized that $(\arctan)' = (1 + x^2)^{-1}$
