Show by direct substitution that $\int_0^{\infty}\frac{1}{1+x^2}\,dx = \int_0^{1}\frac{1}{\sqrt{1-t^2}}\,dt$

Question

I have the following problem: Show by direct substitution ($t=x/\sqrt{1+x^2}$) that $\int_0^{\infty}1/(1+x^2)dx = \int_0^{1}1/\sqrt{1-t^2}dt$.

This is a dumb question but I have not done this in a while and it has take me quite some time to do it. I feel there is a clean way to do it but every approach I take just gets messy. I can rearrange the substitution to obtain $x^2=\frac{t^2}{1-t^2}$, and then $1+x^2=\frac{1}{1-t^2}$. But any attempt at substituting this in just gives me $1-t^2$ in the numerator. The limits I am able to get by just looking at $t=x/\sqrt{1+x^2}$ going to the limits. When calculating $dt = \frac{\frac{1-x^2}{\sqrt{1+x^2}}}{1+x^2}$. But this just gets nasty.

Please enlighten me.

Thanks in advance!

Answer 1

Your approach is correct, but when calculating $\mathrm{d}t$ you made a mistake. Since $$t=\frac{x}{\sqrt{1+x^2}} $$ You have that $$\mathrm{d}t=\frac{\sqrt{1+x^2}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}\mathrm{d}x=\frac{\mathrm{d}x}{(1+x^2)\sqrt{1+x^2}} $$ Since $$x^2=\frac{t^2}{1-t^2}$$ $$\sqrt{1+x^2}=\frac{1}{\sqrt{1-t^2}}$$ Finally $$\int_0^{+\infty}\frac{1}{1+x^2}\mathrm{d}x=\int_0^1\frac{\mathrm{d}t}{\sqrt{1-t^2}}$$

Answer 2

Use subtitution: $x=\tan\arcsin t$.

In addition, Just notice its indefinite integral, You can easily come to this subtitution.($\arctan x=\arcsin t$)

This answer is equivalent to other answers，because:

$$\tan x =\frac{\sin x}{\cos x}= \frac{\sin x}{\sqrt[]{1-\sin ^2 x} }\quad (0<x<\frac{\pi}{2} )$$

We have $\tan \arcsin t = \frac{t}{\sqrt[]{1-t^2} } $.

Answer 3

If $$t=\frac{x}{\sqrt{1+x^2}},$$ than $$\textrm{d}t = \frac{1}{(1+x^2)^{3/2}}\;\textrm{d}x.$$ Thus, this gives: $$\int_0^1\frac{1}{\sqrt{1-t^2}}\;\textrm{d}t = \int_0^\infty \frac{1}{\sqrt{1-\left(\cfrac{x}{\sqrt{1+x^2}}\right)^2}}\cdot \frac{1}{(1+x^2)^{3/2}}\;\textrm{d}x$$ Which finaly gives: $$\int_0^\infty \sqrt{1+x^2}\cdot \frac{1}{(1+x^2)^{3/2}}\;\textrm{d}x = \int_0^\infty\frac{1}{1+x^2}\;\textrm{d}x$$