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

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

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

The value of int_((pi)/(6))^((pi)/(3)) (dx)/(1+sqrt(tan x)) is equal to -

int_((-pi)/(4))^((pi)/(4)) (dx)/(1+e^(tan x))

int_((pi)/(6))^((pi)/(3))(dx)/(1+sqrt(tanx))=(pi)/(12)

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

I_(1) = int_(pi/6)^(pi/3) (dx)/(1+sqrt(tanx)) and I_(2) = (sqrt(sinx)dx)/(sqrt(sinx) + sqrt(cosx)) What is I_(1) - I_(2) equal to ?

I_(1) = int_(pi/6)^(pi/3) (dx)/(1+sqrt(tanx)) and I_(2) = (sqrt(sinx)dx)/(sqrt(sinx) + sqrt(cosx)) What is I_(1) equal to ?

Prove that int_(a)^(b)f(x)dx=int_(a)^(b)f(a+b-x)dx and hence evaluate int_((pi)/(6))^((pi)/(3))(1)/(1+sqrt(tanx))dx.