Question

If $I={\int }_0^{\frac{\pi }{2}}\log \left(\cos x\cos 2nx\right)dx$ then $1$ is same as

• A. ${\int }_0^{\frac{\pi }{2}}\cot x\cos \left(2nx\right)dx$
• B. ${\int }_0^{\frac{\pi }{2}}\cot x\sin \left(2nx\right)dx$
• C. ${\int }_0^{\frac{\pi }{2}}\tan x\cos \left(2nx\right)dx$
• D. $\frac{1}{2n}{\int }_0^{\frac{\pi }{2}}\tan x\sin \left(2nx\right)dx$

Answer 1

The correct option is D $\frac{1}{2n}{\int }_0^{\frac{\pi }{2}}\tan x\sin \left(2nx\right)dx$

$I={\int }_0^{\frac{\pi }{2}}\log \left(\cos x\cos 2nx\right)dx$
$I={\int }_0^{\frac{\pi }{2}}\log \left(\sin x\sin 2nx\right)dx$
Adding the two integrals
$2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin x\cos x\cos 2nx\sin 2nx\right)dx$
$2I={\int }_0^{\frac{\pi }{2}}\log \left(\sin 2x\sin 4nx\right)dx-{\int }_0^{\frac{\pi }{2}}\log 4dx$
$2I={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx+{\int }_0^{\frac{\pi }{2}}\log \sin 4nxdx-\frac{\pi }{2}\log 4$
$I_1={\int }_0^{\frac{\pi }{2}}\log \sin 2xdx$
$I_1={\int }_0^{\frac{\pi }{2}}\log \cos 2xdx$
${2I}_1={\int }_0^{\frac{\pi }{2}}\log \sin 2x\cos 2xdx$
${2I}_1={\int }_0^{\frac{\pi }{2}}\log \sin 4xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx$
Put $2x=t$
When $x=0,t=0$ and when $x=\frac{\pi }{2},t=\pi$
${2I}_1=\frac{1}{2}{\int }_0^{\pi }\log \sin 2tdt-\frac{\pi }{2}\log 2$
${2I}_1={\int }_0^{\frac{\pi }{2}}\log \sin 2tdt-\frac{\pi }{2}\log 2$
$I_1=-\frac{\pi }{2}\log 2$
$I_2={\int }_0^{\frac{\pi }{2}}\log \sin 4nxdx$
$I_2={\int }_0^{\frac{\pi }{2}}\log \cos 4nxdx$
$2I_2={\int }_0^{\frac{\pi }{2}}\log \sin 4n.2xdx-{\int }_0^{\frac{\pi }{2}}\log 2dx$
Put $2x=u$
When $x=0,u=0$ and when $x=\frac{\pi }{2},u=\pi$
$2I_2=\frac{1}{2}{\int }_0^{\pi }\log \sin 4nudu-\frac{\pi }{2}\log 2$
${2I}_2={\int }_0^{\frac{\pi }{2}}\log \sin 4nudu-\frac{\pi }{2}\log 2$
$\Rightarrow I_2=-\frac{\pi }{2}\log 2$
So, $2I=-2\pi \log 2$
$I=-\pi \log 2$