Question

If $\int \frac{1+\cos x}{\cos x-{\cos }^{2}x}dx=\log |\sec x+\tan x|-2f^{\prime }\left(x\right)+C$, then the function $f\left(x\right)$ is

• A. $2\cot \left(\frac{x}{2}\right)$
• B. $2\ln \left|\sin \frac{x}{2}\right|$
• C. $-2\cot \left(\frac{x}{2}\right)$
• D. $-2\ln \left|\sin \frac{x}{2}\right|$

Answer 1

The correct option is B $2\ln \left|\sin \frac{x}{2}\right|$

$\int \frac{1+\cos x}{\cos x-{\cos }^{2}x}dx=\log |\sec x+\tan x|-2f^{\prime }\left(x\right)+C$

$\int \frac{1+\cos x}{\cos x-{\cos }^{2}x}dx=\int \frac{1-\cos x+2\cos x}{\cos x(1-\cos x)}dx$
$=\int \frac{1}{\cos x}dx+\int \frac{2}{1-\cos x}dx$
$=\int \sec xdx+\int {cosec}^2\left(\frac{x}{2}\right)dx$
$=\ln |\sec x+\tan x|-2\cot \left(\frac{x}{2}\right)+C$
$\Rightarrow -2f^{\prime }\left(x\right)=-2\cot \left(\frac{x}{2}\right)$
$f\left(x\right)=\int \cot \left(\frac{x}{2}\right)dx$
$\therefore f\left(x\right)=2\ln \left|\sin \frac{x}{2}\right|$