Question

$\int \frac{1}{\sec x+co\sec x}dx=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2\sqrt{2}}\log \tan (\frac{x}{2}+\frac{\pi }{2k})$. Find the value of $k$.

Answer 1

Let $I=\int \frac{1}{\sec x+co\sec x}dx=\int \frac{\sin x\cos x}{(\sin x+\cos x)}dx$
$=\frac{1}{2}\int \frac{\sin 2xdx}{(\sin x+\cos x)}=\frac{1}{2}\int \frac{{(\sin x+\cos x)}^2-1}{\sin x+\cos x}dx$
$=\frac{1}{2}\int (\sin x+\cos x)dx-\frac{1}{2}\int \frac{dx}{\sin x+\cos x}$
$=\frac{1}{2}[-\cos x+\sin x]-\frac{1}{2\sqrt{2}}\log \tan (\frac{x}{2}+\frac{\pi }{8})$