MathematicsHard64×since 2004Q4010Let a∈(0,π2)a \in \left( {0,{\pi \over 2}} \right)a∈(0,2π) be fixed. If the integral ∫tanx+tanαtanx−tanαdx\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx∫tanx−tanαtanx+tanαdx = A(x) cos 2α\alphaα + B(x) sin 2α\alphaα + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively :Ax−αx - \alphax−α and loge∣cos(x−α)∣{\log _e}\left| {\cos \left( {x - \alpha } \right)} \right|loge∣cos(x−α)∣Bx+αx + \alphax+α and loge∣sin(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|loge∣sin(x−α)∣Cx+αx + \alphax+α and loge∣sin(x+α)∣{\log _e}\left| {\sin \left( {x + \alpha } \right)} \right|loge∣sin(x+α)∣Dx−αx - \alphax−α and loge∣sin(x−α)∣{\log _e}\left| {\sin \left( {x - \alpha } \right)} \right|loge∣sin(x−α)∣Check answerSkip