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.

Options

A. `x+alpha "and log"_(e)|"sin"(x+alpha)|`
B. `x-alpha "and log"_(e)|"sin"(x-alpha)|`
C. `x-alpha "and log"_(e)|"cos"(x-alpha)|`
D. `x+alpha "and log"_(e)|"sin"(x-alpha)|`

Answer

B

Explanation

Let `I = int(tan x + tan alpha)/(tan x - tan alpha)dx, alpha in (0,(pi)/(2))`
`= int((sinx)/(cosx)+(sin alpha)/(cos alpha))/((sinx)/(cosx)-(sin alpha)/(cos alpha))dx`
`=int(sin x cos alpha + sin alpha cos x)/(sin x cos alpha - sin alpha cos x)dx`
`=int(sin(x+alpha))/(sin(x-alpha))dx`
Now, put `x-alpha = t rArr dx = dt, so`
`I=int(sin(t+2 alpha))/(sin t)dt`
`=int(sin t cos 2 alpha + sin 2 alpha cos t)/(sin t)dt`
`=int(cos 2 alpha + sin 2 alpha (cos t)/(sin t))dt`
`= t(cos 2 alpha)+(sin 2 alpha)log_(e)|sin t|+C`
`=(x-alpha) cos 2 alpha +(sin 2 alpha)log_(e)|sin(x-alpha)|+C`
`= A(x) cos 2 alpha + B(x) sin 2 alpha + C` (given)
Now on comparing, we get
`A(x) = x - alpha and B(x) = log_(e)|sin(x-alpha)|`