(a)OR(b)

Find: $\displaystyle \int \frac{x+\sin x}{1+\cos x} d x$

$\displaystyle \int \frac{x+\sin x}{1+\cos x}\,dx$. Using $\displaystyle 1+\cos x = 2\cos^2\frac{x}{2}$ and $\displaystyle \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}$: the integrand $\displaystyle = \frac{x}{2}\sec^2\frac{x}{2} + \tan\frac{x}{2}$. Integrating the first term by parts against $\displaystyle \int \tan\frac{x}{2}$ cancels the $\displaystyle \tan$ terms, giving $\displaystyle \boxed{x\tan\frac{x}{2} + C}$.