Question
What is $$\int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\cot \,x} \right)dx} $$ equal to ?
A.
$$0$$
B.
$$\pi \,\ell n\,2$$
C.
$$ - \pi \,\ell n\,2$$
D.
$$\frac{{\pi \,\ell n\,2}}{2}$$
Answer :
$$0$$
Solution :
$$\eqalign{
& {\text{Let }}I = \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\cot \,x} \right)dx} \cr
& = \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\cos \,x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr
& = \int\limits_0^{\frac{\pi }{2}} {\sin \,\left[ {2\left( {\frac{\pi }{2} + x} \right)} \right]\ell n\,\cos \left( {\frac{\pi }{2} + x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr
& = \int\limits_0^{\frac{\pi }{2}} {\sin \left( {\pi + 2x} \right)\ell n\left( {\sin \,x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr
& = \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} - \int\limits_0^{\frac{\pi }{2}} {\sin \,2x\,\ell n\left( {\sin \,x} \right)dx} \cr
& = 0 \cr} $$