Question

Solve $\int \frac{dx}{1+\sin x}$

Answer 1

Let $I=\int \frac{dx}{1+\sin x}$
As $1+\sin x={(\cos \frac{x}{2}+\sin \frac{x}{2})}^2$
$=2{(\frac{1}{\sqrt{\left(2\right)}}\cos \frac{x}{2}+\frac{1}{\sqrt{\left(2\right)}}\sin \frac{x}{2})}^2=2{\cos }^2(\frac{x}{2}-\frac{\pi }{4})$
Therefore
$I=\frac{1}{2}\int {\sec }^2(\frac{x}{2}-\frac{\pi }{4})dx=\frac{1}{2}.2\tan (\frac{x}{2}-\frac{\pi }{4})+c$
$=\tan (\frac{x}{2}-\frac{\pi }{4})+b$
Therefore $a=-\pi /4,b$ is any constant