Question

Solve ${\int }_{\pi /6}^{\pi /3}\frac{1}{1+\tan x}dx$

Answer 1

$I={\int }_{\pi /6}^{\pi /3}\frac{1}{1+\tan x}dx$

$\Rightarrow I={\int }_{\pi /6}^{\pi /3}\frac{1}{1+\frac{\sin x}{\cos x}}dx$

$I={\int }_{\pi /6}^{\pi /3}\frac{\cos x}{\sin x+\cos x}dx...\left(1\right)$

$I={\int }_{\pi /6}^{\pi /3}\frac{\cos x\left(\frac{\pi }{3}+\frac{\pi }{6}-x\right)dx}{\sin x\left(\frac{\pi }{3}+\frac{\pi }{6}-x\right)+\cos x\left(\frac{\pi }{3}+\frac{\pi }{6}-x\right)}$

$I={\int }_{\pi /6}^{\pi /3}\frac{\sin x}{\cos x+\sin x}dx...\left(2\right)$

Adding equation (1) and (2)

$2I={\int }_{\pi /6}^{\pi /3}\frac{sinx+\cos x}{\sin x+\cos x}dx$

$2I={{\left[x\right]}_{\pi /6}}^{\pi /3}$

$2I=\frac{\pi }{3}-\frac{\pi }{6}$

$2I=\frac{\pi }{6}$

$I=\frac{\pi }{12}$