Easiest way to solve this integral

Question

I was solving this problem from a calculus textbook and I got stuck at this particular problem. I tried to put it into Integral Calculator after I was unable to solve it, but now I wonder if there is an easier way.

What is the easiest way to solve the following indefinite integral: $$\int \frac{x dx}{1 + \cos x}, x \in (-\pi, \pi)$$

Thank you very much.

Answer 1

Use $$\cos x=2\cos^2\frac x2-1,$$ then \begin{align*} \int\frac x{1+\cos x}dx&=\int\frac x{2\cos^2 \frac x2}dx\\[3pt] &=\int x\sec^2\frac x2\cdot \frac12dx\\[3pt] &=\int x\cdot d\left(\tan \frac x2\right)\\[3pt] &=x\tan \frac x2-\int\tan \frac x2dx\tag{P}\\[3pt] &=x\tan \frac x2+2\log\left(\cos \frac x2\right)+C \end{align*} Where (P) is integration by parts.

Answer 2

Multiplying by $\displaystyle\frac{1-\cos x}{1-\cos x}$ gives $\displaystyle\int\frac{x(1-\cos x)}{\sin^2x}dx=\int x\left(\csc^2x -\csc x\cot x\right) dx$.

Now integrate by parts, with $u=x$ and $dv=(\csc^2x-\csc x \cot x) dx$ and $v=\csc x-\cot x$ to get

$\hspace{.3 in}\displaystyle x(\csc x-\cot x)+\ln|\sin x|-\ln|\csc x-\cot x|+C$

$\hspace{.3 in}\displaystyle=\frac{x\sin x}{1+\cos x}+\ln(1+\cos x)+C$