Evaluate $\int \frac{2 \sin x+ \sin 2x}{1 - \cos x }\mathrm{d}x $ by substitution

Question

$$\int \frac{2 \sin x+ \sin 2x}{1 - \cos x }\mathrm{d}x . $$

the following to is to be solved by substitution method. I tried to substitute $1-\cos x$, but couldn't reach the solution. Any help is appreciated.

Answer 1

$$\begin{align}\int \frac{2 \sin x+ \sin 2x}{1 - \cos x }\ dx &= \int \frac{2 \sin x+ 2\sin x\cos x}{1 - \cos x }\ dx \\ &= 2\int\frac{1+\cos x}{1-\cos x}\sin x\ dx \\ &= 2\int\frac{-(1-\cos x)+2}{1-\cos x}\frac{d(1-\cos x)}{dx}\ dx \\ &= 2\int\frac{-(1-\cos x)+2}{1-\cos x}\ {d(1-\cos x)}\end{align}$$

Does that help?

Answer 2

Hint

$$I=\int \frac{2 \sin( x)+ \sin (2x)}{1 - \cos (x) }\ dx = \int \frac{2 \sin (x)+ 2\sin ( x)\cos ( x)}{1 - \cos ( x) }\ dx = 2\int\frac{1+\cos( x)}{1-\cos( x)}\sin ( x)\ dx $$

Now, change variable $y=\cos(x)$, $dy=-\sin(x)\,dx$ which makes $$I=-2\int\frac{1+y}{1-y}\,dy$$ which seems easy to solve.

Answer 3

$$u=1-\cos t$$ $$\int \frac{2 \sin x+ 2\sin x \cos x}{1 - \cos x }\ dx=2\int {2-u \over u} du$$

$$=\int (\frac4u-2) du=-2u+4\log u +C$$ $$=-2(1-\cos t)+4\log (1-\cos t) +C$$ $$= 2 \cos t +4\log (1-\cos t) +C'$$