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∫ cosec x / log(tan x/2) dx , please ans asap, thank you.
This can be written as1/sin x* log tan(x/2)= 1/ 2sin(x/2) cos(x/2) * log tan(x/2)Multiply by cos (x/2) in both numerator or denominator= 1/ 2tan(x/2) cos²(x/2) * log tan(x/2) Now assume Log tan(x/2) = tSec²(x/2)/ 2*tan(x/2) dx = dtdx/ 2 cos²(x/2) tan(x/2) = dtHope now you can do it further
Using substitution , denote the denominator by tt=log tanx/2 , differentiating on both sides dt/dx = 1/tanx/2×sec^2 x/2×1/2Hence we get dt =1/sinx dx Substitute the value of dx as dt×sinx in cosec x/ t dtFrom this we get log log tan x/2 + c
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