Question

Solve:
${\int }_{\frac{\pi }{6}}^{\frac{\pi }{3}}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx$

Answer 1

$I{\int }_{\pi /6}^{\pi /3}\frac{sinx+cosx}{\sqrt{2sinx+cosx}}dx={\int }_{\pi /6}^{\pi /3}\frac{1}{\sqrt{2}}\sqrt{\frac{sinx}{cosx}}dx+{\int }_{\pi /6}^{\pi /3}\frac{1}{\sqrt{2}}\sqrt{\frac{cosx}{sinx}}dx$

Put $tanx=t^2$

$si{n}^{2}dx=2tdt$

$dx=\frac{2t}{1+t^2}dt$

$\Rightarrow {\int }_{\sqrt{\frac{1}{\sqrt{3}}}}^{\sqrt{\sqrt{3}}}\frac{1}{\sqrt{2}}t\to \frac{2t}{(1+t^2)}dt+\frac{1}{\sqrt{2}}\times \frac{2t}{t(1+t^2)}dt$

$\Rightarrow {\int }_{\sqrt{\frac{1}{2}}}^{\sqrt{3}}\frac{1}{\sqrt{2}}\left[\frac{2t^2+2}{(t^2+1)}\right]dt$

$\Rightarrow \sqrt{2}\times {\int }_{\sqrt{\frac{1}{\sqrt{3}}}}^{\sqrt{3}}dt$

$\Rightarrow \sqrt{2}+{]}_{\sqrt{\frac{1}{\sqrt{3}}}}^{\sqrt{3}}$

$I\Rightarrow \sqrt{2}[3^{1/4}-3^{-1/4}]$