There are many formulas for π .
One of most known formulas is
\[ \int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\pi, \qquad(-\infty\lt x\lt \infty) . \]
Please put $x=f(t)=\tan t=\frac{\sin t}{\cos t},\quad (t\in (-\pi/2,\pi/2))$ .
As we already have known
\[ \cos't=-\sin t,\quad \sin' t=\cos t , \]
we will obtain
\[ \frac{dx}{dt}=\frac{df(t)}{dt}=f'(t)=\frac{\cos^2 t+\sin^2 t}{\cos^2 t}=\frac{1}{\cos^2 t} . \]
Then,
\[ f'(t)=1+f^2(t), \quad\mbox{and}\quad dx=f'(t)dt \]
$f$ is a monotonically increasing mapping of $(-\pi/2,\pi/2)$ onto $(-\infty,\infty)$.
Thus,
\[ \int_{-\infty}^{\infty}\frac{dx}{1+x^2}=\int_{-\pi/2}^{\pi/2}\frac{f'(t)dt}{1+f^2(t)}=\int_{-\pi/2}^{\pi/2}dt=\pi \]
