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Find the value of int0^(pi/2)sin2xlogta...

Find the value of `int_0^(pi/2)sin2xlogtanxdx`

Text Solution

Verified by Experts

The correct Answer is:
`0`
`I=int_(0)^(pi//2)sin2x log tan x dx`
`=int_(0)^(pi//2) sin2((pi)/2-x)"log tan"((pi)/2-x)dx`
`=-int_(0)^(pi//2)sin 2x log tan x dx=-1`
or `2I=0`
or `I=0`
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