Question

Prove that ${\int }_0^{\pi /2}\frac{dx}{1+\tan x}=\frac{\pi }{4}$.

Answer 1

$I={\int }_0^{\pi /2}\frac{dx}{1+\tan x}$

$={\int }_0^{\pi /2}\frac{dx}{1+\frac{\sin x}{\cos x}}$

$={\int }_0^{\pi /2}\frac{\cos x}{\cos x+\sin x}dx$ ..........(1)

$={\int }_0^{\pi /2}\frac{\cos (\pi /2-x)}{\cos (\pi /2-x)+\sin (\pi /2-x)}dx$

$[$By property ${\int }_0^{a}f\left(x\right)dx={\int }_0^{a}f(a-x)dx]$

$I={\int }_0^{\pi /2}\frac{\sin x}{\sin x+\cos x}dx$ ........(2)

Adding equation (1) and (2) $I+I={\int }_0^{\pi /2}\frac{\sin x+\cos x}{\sin x+\cos x}dx$

$\Rightarrow 2I={\int }_0^{\pi /2}1dx$

$\Rightarrow 2I=[x{]}_0^{\pi /2}$

$\Rightarrow 2I=\frac{\pi }{2}-0$

$\Rightarrow I=\frac{\pi }{4}$.