Question

$\int(\cot \,x \cot (x+\alpha)+1) d x=$

• 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)

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