Question

Integrate :
$\sqrt{\frac{1+sinx}{1-sinx}}$

Answer 1

$I=\int \sqrt{\frac{1+\sin x}{1-\sin x}}dx=\int \sqrt{\frac{(1+\sin x)(1-\sin x)}{(1-\sin x{)}^2}}dx=\int \sqrt{\frac{1-{\sin }^{2}x}{(1-\sin x{)}^2}}dx$

$=\int \frac{\cos x}{1-\sin x}dx$

$1-\sin x=t$

$-\cos xdx=dt$

$\Rightarrow -\int \frac{dt}{t}$

$\Rightarrow -ln|t|+c$

$\Rightarrow -ln|1-\sin x|+c$

$\Rightarrow ln\left|\frac{1}{1-\sin x}\right|+c$

$\therefore \int \sqrt{\frac{1+\sin x}{1-\sin x}}=ln\left|\frac{1}{1-\sin x}\right|+c$.