$\underset{0}{\overset{4}{\int }}\frac{\mathit{d}\mathit{x}}{1+{\mathit{x}}^2}$ dx
Let x = tan θ ⇒ θ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ x
dx = $\mathit{s}\mathit{e}{\mathit{c}}^2$ θ d θ
When x = 0 , θ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ (0) = 0
When x = 1, θ = $\mathit{t}\mathit{a}{\mathit{n}}^{-1}$ 1 = $\frac{\mathit{\pi }}{4}$
∴ $\underset{0}{\overset{4}{\int }}\frac{\mathit{d}\mathit{x}}{1+{\mathit{x}}^2}$ = $\underset{0}{\overset{\frac{\mathit{\pi }}{4}}{\int }}$ $\frac{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }}{1+\mathit{t}\mathit{a}{\mathit{n}}^2\mathit{\theta }}$ dθ
= $\underset{0}{\overset{\frac{\mathit{\pi }}{4}}{\int }}\frac{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }}{\mathit{s}\mathit{e}{\mathit{c}}^2\mathit{\theta }}$ dθ
= $\underset{0}{\overset{\frac{\mathit{\pi }}{4}}{\int }}$ dθ
= ${\left[\mathit{\theta }\right]}_0^{\frac{\mathit{\pi }}{4}}$
= $\left[\frac{\mathit{\pi }}{4}-0\right]$
= $\frac{\mathit{\pi }}{4}$