If `int tan(x-alpha)tan(x+alpha)*tan2xdx=plog|sec2x|+qlog|sec(x+alpha)|+rlog|sec(x-alpha)|+c` then p+q+r= . . ..

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 alpha)/(tan x+tan alpha)dx

int (tan x - tan alpha)/(tan x+tan alpha)dx

tan^(2)alpha=1-p^(2) then sec alpha+tan^(3)alpha*cos ec alpha

Evaluate: inttan(x+alpha)tan(2x-alpha)tan3xdx

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.

int sec^(p)x tan xdx