Question

$\int \frac{\cos \alpha }{\sin x\cos (x-\alpha )}dx=$ ____ $+c$ where $0<x<\alpha <\frac{\pi }{2}$ and $\alpha$ is a constant.

• A. $-\log |\cot x+\tan \alpha |$
• B. $\log |\cot x+\tan \alpha |$
• C. $-\log |\tan x+\cot \alpha |$
• D. $\log |\tan x+\cot \alpha |$

Answer 1

The correct option is A $-\log |\cot x+\tan \alpha |$

$\int \frac{\cos (x-(x-\alpha )}{\sin x\cos (x-\alpha )}dx=\int \frac{\cos (x-\alpha )\cos x+\sin (x-\alpha )\sin x}{\sin x\cos (x-\alpha )}dx$
$=\int (\cot x+\tan (x-\alpha ))dx$

$=\ln |\sin x|-\ln |\cos (x-\alpha )|+C=-\ln \left|\frac{\cos (x-\alpha )}{\sin x}\right|+C=-\ln \left|\frac{\cos \alpha \cos x+\sin \alpha \sin x}{\sin x}\right|+C$

$=-\ln |\cos \alpha \cot x+\sin \alpha |+C$

$=-\ln \left|\cos \alpha \right(\cot x+\tan \alpha \left)\right|+C$

$=-\ln \left|\right(\cot x+\tan \alpha \left)\right|+c$,
where $c=-\ln \cos \alpha +C$