`int_0^(pi//2)sin2xlogtanx\ dx\ ` is equal to `pi//2` b. `pi` c. `0` d. `2pi`

Explanation

`0`


`I=int_{0}^{frac{pi}{2}} sin 2 x log tan x d x ldots ldots(1) `

`I=int_{0}^{frac{pi}{2}} sin (pi-2 x) log tan (frac{pi}{2}-x) d x `

`I=int_{0}^{frac{pi}{2}} sin 2 x log cot x d x ldots . .(2)`


Adding `(1)` and `(2)`, we get,

`2 I=int_{0}^{frac{pi}{2}} sin 2 x(log tan x+log cot x) d x `

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