Question

Integrate (x+sinx)/(1+cosx)

Answer 1

maybe expanding it would help

x sin x
-------- + ---------
1 + cos x 1 + cos x

Answer 2

- just a thought...

Answer 3

Let \(u=\tan\dfrac{x}{2}\), or \(x=2\arctan u\), so that \(dx=\dfrac{2}{1+u^2}\,du\).

Since \(\tan\dfrac{x}{2}=u\), it follows that \(\sin\dfrac{x}{2}=\dfrac{u}{\sqrt{1+u^2}}\) and \(\cos\dfrac{x}{2}=\dfrac{1}{\sqrt{1+u^2}}\).

Recall some identities:
\[\begin{align*}\sin x&=\sin\left(2\times\frac{x}{2}\right)\\[1ex]&=2\sin\frac{x}{2}\cos\frac{x}{2}\\[1ex]&=\frac{2u}{1+u^2}\\[2ex]
\cos x&=\cos\left(2\times\frac{x}{2}\right)\\[1ex]&=\cos^2\frac{x}{2}-\sin^2\frac{x}{2}\\[1ex]&=\frac{1-u^2}{1+u^2}\end{align*}\]
All this to say that
\[\int\frac{x+\sin x}{1+\cos x}\,dx=\int\frac{2\arctan u+\dfrac{2u}{1+u^2}}{1+\dfrac{1-u^2}{1+u^2}}\times\frac{2}{1+u^2}\,du\]
We're in luck - this simplifies nicely:
\[4\int\left(\arctan u+\dfrac{u}{1+u^2}\right)\,du\]