How to integrate $\frac{\cos x-\cos2x}{1-\cos x}$? I want to find $$\int\dfrac{\cos x-\cos2x}{1-\cos x}\ dx$$
I have tried solving the question using substitution. how do I solve it? 

Notice, we have $$\int\frac{\cos x-\cos 2x}{1-\cos x}dx$$
$$=\int\frac{\cos x-2\cos^2x+1}{1-\cos x}dx$$
$$=\int\frac{\cos x-2\cos^2x+1}{1-\cos x}dx$$
$$=\int\frac{\cos x-\cos^2x+1-\cos^2 x}{1-\cos x}dx$$
$$=\int\frac{\cos x(1-\cos x)+(1-\cos^2 x)}{1-\cos x}dx$$
$$=\int\frac{\cos x(1-\cos x)}{1-\cos x}dx+\int\frac{(1+\cos x)(1-\cos x)}{1-\cos x}dx$$
$$=\int \cos x dx+\int (1+\cos x)dx$$
$$= \sin x +x+\sin x+C$$
$$= 2\sin x+x+C$$
