Question

Evaluate : $\int \frac{1+2x\sin x-\cos x}{x(1-\cos x)}dx.$

• A. $\log x+2\log (1-\cos x)$.
• B. $\log x-2\log (1-\cos x)$.
• C. $\log x+2\log (1+\cos x)$.
• D. $\log x+2\log (1-\sin x)$.

Answer 1

The correct option is A $\log x+2\log (1-\cos x)$.

$\int \frac{1+2xsinx-cosx}{x(1-cosx)}dx$

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

$=\int \frac{1}{x}dx$ + $\int \frac{2sinx}{1-cosx}dx$

$=\log x+\int \frac{2sinx}{1-cosx}dx$

Let, $-cosx$ be $t,$ Then $dt=d(-cosx)=sinx$

$=\log x+\int \frac{2}{1+t}dt$

$=\log x+2\log (1+t)$

$=\log x+2\log (1-cosx)$