`int{1 + tan x.tan(x+alpha)}dx= `

int[1+tanx.tan(x+alpha)]dx is equal to

int[1+tanx.tan(x+alpha)]dx is equal to

Evaluate: int{1+tan x tan(x+theta)}dx

The value of int {1+tan x*tan (x + A)}\ dx is

The value of int {1+tan x*tan (x + A)}\ dx is

int tan(x-alpha).tan(x+alpha).tan 2x dx is equal to

int tan(x-alpha).tan(x+alpha).tan 2x dx is equal to

int tan x tan 2x tan 3x dx

Let alpha in (0, pi/2) be fixed. If the integral int(tan x + tan alpha)/(tan x - tan alpha)dx = A(x)cos 2 alpha+B(x) sin 2 alpha +C ,where C is a constant of integration, then the functions A(x) and B(x) are respectively.