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" (T) "ftag(sin pi)/(1+tan)-int(0)^( pi/...

" (T) "ftag(sin pi)/(1+tan)-int_(0)^( pi/2)(dx)/(1+sqrt(tan x))=(pi)/(4)

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int_(0)^(pi//2)(1)/(1+sqrt(tan x))dx=

int_(0)^( pi/2)(dx)/(1+tan^(5)x)

int_(0)^(pi//2)(dx)/(1+tan^(3)x)=

int_(pi//6)^(pi//3) (dx)/(1 + sqrt(tan x)) =

int_(0)^((pi)/(2))(1)/(1+tan x)dx

int_(0)^((pi)/(2))(dx)/(1+sqrt(tan x))=int_(0)^((pi)/(2))(dx)/(1+sqrt(cot x))=(pi)/(4)

Evaluate :int_(0)^((pi)/(2))(dx)/(1+sqrt(tan x))