Question

$\int \frac{1}{\sqrt{(1+\sin x)}}dx=\sqrt{2}\log \tan (x/4+\pi /4k).$ Find the value of $k$.

Answer 1

$\int \frac{dx}{\sqrt{(1+\sin x)}}=\int \frac{dx}{\sqrt{({\sin }^2\left(x/2\right)+{\cos }^2\left(x/2\right)+2\sin \left(x/2\right)\cos \left(x/2\right))}}$
$=\int \frac{dx}{\sin \frac{x}{2}+\cos \frac{x}{2}}=\int \frac{1}{\sqrt{2}(\frac{1}{\sqrt{\left(2\right)}}\sin \frac{x}{2}+\frac{1}{\sqrt{\left(2\right)}}\cos \frac{x}{2})}dx$
$=\int \frac{1}{\sqrt{\left(2\right)}\sin (x/2+\pi /4)}dx=\frac{1}{\sqrt{\left(2\right)}}\int co\sec (x/2+\pi /4)dx$
$=\left(1/\sqrt{2}\right).2\log \tan (x/4+\pi /8)=\sqrt{2}\log \tan (x/4+\pi /8)$