Question

Evaluate: ${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}$

Answer 1

Let $I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{dx}{1+\sqrt{\tan x}}={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}dx$........(1)

Now using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$

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

Now add (1) and (2)

$2I={\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}dx=\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$

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