Discussion Forum

$\int_{0}^{\pi/2} \sin 2x.\log \tan x dx$ is equal to

Answer:

Explanation:

Let   $I=\int_{0}^{\pi/2} (\log \tan x).\sin 2 x dx$   ......(i)

$I=\int_{0}^{\pi/2} \log \tan(\frac{\pi}{2}- x).\sin 2(\frac{\pi}{2}- x) dx$

                                           $\left[\because\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx\right]$

$\Rightarrow I=\int_{0}^{\pi/2} \log \cot x.\sin 2 x dx$...........(ii)

                                         $[\because\sin (\pi-2x)=\sin 2x]$

 On adding eqs(i) and (ii) , we get

$2I\int_{0}^{\pi/2} \log\tan x.\sin2x dx+\int_{0}^{\pi/2} \log\cot x\sin 2x dx$

=  $\int_{0}^{\pi/2} \sin2x\log(\tan x.\cot x) dx$

$[\because$   $\log m+\log n=\log (m.n)]$

 $\int_{0}^{\pi/2} \sin 2x\log1 dx$

$\Rightarrow$   $1=0[\because log 1=0]$

$\therefore$   $\int_{0}^{\pi/2} \sin 2x\log (\tan x) dx=0$