Question

The value of $\int \frac{\cos x-\cos 2x}{1-\cos x}dx$ equals?

• A. $x+\sin x+c$
• B. $x-2\sin x+c$
• C. $-x+2\cos x+c$
• D. $x-2\cos x+c$

Answer 1

The correct option is A $x+\sin x+c$

$\Rightarrow \int \frac{\cos x-\cos 2x}{\left(1-\cos x\right)}$

$\Rightarrow \int \frac{\left(1+\cos x\right)-1-\cos 2x}{\left(1-\cos x\right)}dx$

$\Rightarrow \int \left[\frac{-\left(1-\cos x\right)}{(1-\cos x}\frac{+1-\cos 2x}{1-\cos x}\right]dx$

$\Rightarrow \int -1dx+\int \frac{2{\sin }^{2}x}{2{\sin }^{2}x/2}dx$

$\Rightarrow -x+\int \frac{4{\sin }^{1}x/2.{\cos }^{2}x/2}{{\sin }^{2}x/2}dx$

$\Rightarrow -x+4\int {\cos }^{2}x/2$

$\Rightarrow -x+4\int \frac{\left(1+\cos x\right)}{2}dx$

$\Rightarrow -x+2\int dx+\int \cos xdx$

$\Rightarrow -x+2x+\sin x+c$

$\Rightarrow x+\sin x+c$