Question

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

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

Answer 1

Answer: (D)

$-ln|\cot x+\tan \alpha |$

$\int \frac{cos\alpha }{sinxcos(x-\alpha )}dx$

$=\int \frac{cos(x-(x-\alpha \left)\right)}{sinxcos(x-\alpha )}dx$

$=\int \frac{cosxcos(x-\alpha )+sinxsin(x-\alpha )}{sinxcos(x-\alpha )}dx$

$=\int (cotx+tan(x-\alpha \left)\right)dx$

$=ln|sinx|-ln|cos(x-\alpha )|$

$=ln|\frac{sinx}{cos(x-\alpha )}|$

$=ln|\frac{sinx}{cosxcos\left(\alpha \right)+sinxsin\left(\alpha \right)}|$

$=ln|\frac{sec\left(\alpha \right)}{cotx+tan\left(\alpha \right)}|$

$=ln|\left(sec\right(\alpha \left)\right)-ln(cotx+tan(\alpha \left)\right)|$

$=-ln|(cotx+tan(\alpha \left)\right)|+C$

Hence, the answer is option (D).