Question

Evaluate $\int \frac{\cos x-\cos 2x}{1-\cos x}dx=$

• A. $2\sin +x+c$
• B. $2\cos x+x+c$
• C. $\frac{2\sin x+\cos x}{\cos x}+c$
• D. $2(\sin x+\cos x)+c$

Answer 1

The correct option is A $2\sin +x+c$

$\int \frac{\cos x-\cos 2x}{1-\cos x}dx$

As we know that,

$\cos 2x=2{\cos }^{2}x-1$

$\therefore \int \frac{\cos x-\cos 2x}{1-\cos x}dx=\int \frac{\cos x-(2{\cos }^{2}x-1)}{1-\cos x}dx$

$=\int \frac{-((2{\cos }^{2}x-1)-\cos x)}{-(\cos x-1)}dx$

$=\int \frac{2{\cos }^{2}x-2\cos x+\cos x-1}{\cos x-1}dx$

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

$=\int (2\cos x+1)dx$

$=\int 2\cos xdx+\int 1dx$

$=2\sin x+x+c$

Hence $\int \frac{\cos x-\cos 2x}{1-\cos x}dx=2\sin x+x+c$.