∫ cosec x / log(tan x/2) dx , please ans asap, thank you.

Question

Arun · Answer

This can be written as 1/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) = t Sec²(x/2)/ 2*tan(x/2) dx = dt dx/ 2 cos²(x/2) tan(x/2) = dt Hope now you can do it further

ASHISH yadav · Answer

Using substitution , denote the denominator by t t=log tanx/2 , differentiating on both sides dt/dx = 1/tanx/2×sec^2 x/2×1/2 Hence we get dt =1/sinx dx Substitute the value of dx as dt×sinx in cosec x/ t dt From this we get log log tan x/2 + c

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∫ cosec x / log(tan x/2) dx , please ans asap, thank you.

∫ cosec x / log(tan x/2) dx , please ans asap, thank you.

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