Question

$\int \frac{\sqrt{tanx}}{sinx.cosx}dx$

Answer 1

$\int \frac{\sqrt{tanx}}{sinx.cosx}dx$

$=\int \frac{\sqrt{sinx/cosx}}{sinx.cosx}dx$

$=\int \frac{1}{\sqrt{sinx}{\left(cosx\right)}^{3/2}}dx$

$=\int \frac{1}{\sqrt{\frac{sinx}{cosx}}\sqrt{cosx}{\left(cosx\right)}^{3/2}}dx$

$=\int \frac{1}{\sqrt{tanx}{\cos }^{2}x}dx$

Let $t=tanx\Rightarrow dt={\sec }^{2}dx\Rightarrow dt=\frac{dx}{{\cos }^{2}x}$

Substituting in the above integral we get :

$=\int \frac{1}{\sqrt{t}}dt$

$=\frac{\sqrt{t}}{1/2}+C$

$=2\sqrt{tanx}+C$