Question

The value of the integral
${\int }_0^{3\alpha }cosec(x-\alpha )cosec(x-2\alpha )dx$, is

• A. $2sec\alpha$ $log\left(\frac{1}{2}cosec\alpha \right)$
• B. $2sec\alpha$ $log\left(\frac{1}{2}sec\alpha \right)$
• C. $2cosec\alpha$ $log\left(sec\alpha \right)$
• D. $2cosec\alpha$ $log\left(\frac{1}{2}sec\alpha \right)$

Answer 1

The correct option is D $2cosec\alpha$ $log\left(\frac{1}{2}sec\alpha \right)$

$I={\int }_0^{3\alpha }\csc (x-\alpha )\csc (x-2\alpha )dx=\frac{1}{\sin \alpha }{\int }_0^{3\alpha }\frac{\sin \alpha dx}{\sin (x-\alpha )\sin (x-2\alpha )}$
$=\frac{1}{\sin \alpha }{\int }_0^{3\alpha }\frac{\sin ((x-\alpha )-(x-2\alpha ))dx}{\sin (x-\alpha )\sin (x-2\alpha )}=\frac{1}{\sin \alpha }{\int }_0^{3\alpha }\left(\cot (x-2\alpha )-\cot (x-\alpha )\right)dx$
$=\frac{1}{\sin \alpha }{\left[\log \frac{\sin (x-2\alpha )}{\sin (x-\alpha )}\right]}_0^{3\alpha }=\frac{1}{\sin \alpha }\left[\log \frac{\sin \left(\alpha \right)}{\sin \left(2\alpha \right)}-\log \frac{\sin \left(2\alpha \right)}{\sin \left(\alpha \right)}\right]$
$=2\csc \alpha \log \left(\frac{1}{2}\sec \alpha \right)$