Question

Prove that ${\int }_0^{\pi /2}\sin 2xlog\left(\tan x\right)dx=0$

Answer 1

Let $I={\int }_0^{\pi /2}\sin 2xlog\left(\tan x\right)dx$

$={\int }_0^{\pi /2}\sin (\pi -2x)log(\tan \frac{\pi }{2}-x)dx$ [Using formula of definite integral]

$={\int }_0^{\pi /2}\sin 2xlog\left(\cot x\right)dx=I$

Now $I+I$

$={\int }_0^{\pi /2}\sin 2x\left\{log\right(\tan x)+\log (\cot x\left)\right\}dx$

$={\int }_0^{\pi /2}\sin 2x\left\{log\right(1\left)\right)\}dx=0$

or, $2I=0$

or, $I=0$.