For what value of a and b, the equation $\int \frac{d x}{1+\sin x}=\tan \left(\frac{x}{2}+a\right)+b$ holds good?

a = $-\frac{\pi}{4}$, b is any arbitrary constant

$\int \frac{d x}{1+\sin x}=\int \frac{d x}{1+\cos \left(\frac{\pi}{2}-x\right)}=\frac{1}{2} \int \sec ^2\left(\frac{\pi}{4}-\frac{x}{2}\right) dx$

$=\frac{1}{2} \frac{\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)}{-1 / 2}+c=-\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)+c=\tan \left(\frac{x}{2}-\frac{\pi}{4}\right)+c$

⇒ a = $-\frac{\pi}{4}$, b is any arbitrary constant

Hence (3) is the correct answer.