Question

Evaluate the integrals
${\int }_0^2\frac{1}{(x^2+1)}dx$

Answer 1

Let $x=tan\theta$

$\Rightarrow dx=se{c}^2\theta d\theta$

$\Rightarrow {\int }_0^2\frac{1}{se{c}^2\theta }se{c}^2\theta d\theta$

$\Rightarrow {\int }_0^2\theta ={\int }_0^{2}ta{n}^{-1}x$

$\Rightarrow ta{n}^{-1}2-0$

$=ta{n}^{-1}2$