How to integrate $\frac{\cos x-\cos2x}{1-\cos x}$?

Question

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?

Answer 1

HINT:

$$\cos2x-\cos x=2\cos^2x-1-\cos x$$

Now $$2c^2-c-1=(2c+1)(c-1)$$

Answer 2

Hint: your integrand is equal to $$1+2\cos(x)$$

Answer 3

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