Question

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

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

Answer 1

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

${\int }_0^{3\alpha }\text{cosec}(x-\alpha )\text{cosec}(x-2\alpha )dx$

$={\int }_0^{3\alpha }\frac{1}{\sin (x-\alpha )\sin (x-2\alpha )}dx$

$\frac{1}{\sin \alpha }{\int }_0^{3\alpha }\frac{\sin \left[\right(x-\alpha )-(x-2\alpha \left)\right]}{\sin (x-\alpha )\sin (x-2\alpha )}dx$

$\Rightarrow \frac{1}{\sin \alpha }{\int }_0^{3\alpha }\left[\cot \right(x-2\alpha )-\cot (x-\alpha \left)\right]dx$

$=\frac{1}{\sin \alpha }\left(\log \right[\sin (x-2\alpha )]{|}_0^{3\alpha }-\log (\sin (x-\alpha )\left){|}_0^{3\alpha }\right)$

$=\frac{1}{\sin \alpha }\left[\log \right(\sin \alpha )-\log (\sin (-2\alpha ))-\log (\sin \left(2\alpha \right))+\log (\sin (-\alpha )\left)\right]$

$=2\frac{1}{\sin \alpha }\log \left|\frac{\sin \alpha }{2\sin \alpha \cos \alpha }\right|$

$=2\text{cosec}\alpha \log \left(\frac{1}{2}\sec \alpha \right)$