Question

$\int (\cot x\cot (x+\alpha )+1)dx=$

• A. $\cot \alpha \log \left(|\frac{\sin x}{\sin (x+\alpha )}\mid \right)+c$
• B. $\log |\sin x\sin (x+\alpha )|+x+c$
• C. $\log |\sin x\cos (x+\alpha )|+x+c$
• D. $\tan \alpha \log \left(\left|\frac{\cos x}{\sin (x+\alpha )}\right|\right)+c$

Answer 1

Answer: (A)

$\int [\cot x\cot (x+a)+1]dx$
$=\int \frac{\cos x\cos (x+\alpha )+\sin x\sin (x+\alpha )}{\sin x\sin (x+\alpha )}dx$
$=\int \frac{\cos \alpha }{\sin x\sin (x+\alpha )}dx=\frac{\cos \alpha }{\sin \alpha }\int \frac{\sin \alpha }{\sin x\sin (x+\alpha )}dx$
$=\cos \alpha \int \frac{\sin [(x+\alpha )-x]}{\sin x\sin (x+\alpha )}dx$
$=\cot \alpha \int \frac{\sin (x+\alpha )\cos x-\cos (x+\alpha )\sin x}{\sin x\sin (x+\alpha )}dx$
$=\cot \alpha \int (\cot x-\cot (x+\alpha ))dx$
$=\cot \alpha [\log |\sin x|-\log \sin (x+\alpha )\mid ]+c$
$=\cot \alpha \log \left|\frac{\sin x}{\sin (x+\alpha )}\right|+c$