Question

Let $\alpha \in$ (0, $\pi$/2) be fixed. If the integral
$\int \frac{tan x + tan \alpha}{tan x - tan \alpha} dx =$
A(x) cos2$\alpha$ + B(x) sin2a + C, where C is a constant of integration, then the functions A(x) and B(x) are respectively:

• A. $x - \alpha \, and \, log_e | cos (x - \alpha)|$
• B. $x + \alpha \, and \, log_e | sin (x - \alpha)|$
• C. $x - \alpha \, and \, log_e | sin (x - \alpha)|$
• D. $x + \alpha \, and \, log_e | six (x + \alpha)|$

Answer 1

Answer: (C)

$\int \frac{tan x + tan \alpha}{tan x - tan \alpha} dx = \int \frac{sin (x+ \alpha}{sin (x - \alpha} dx$
Let x - $\alpha = t$
$\Rightarrow \, \, \int \frac{sin (t + 2 \alpha)}{sin t} dt = \int cos2 \alpha dt + \int cot (t) sin 2 \alpha dt$
$= t. cos 2 \alpha + \ell n |sin t|. sin 2 \alpha + C$
$= (x - \alpha) cos2 \alpha + in |sin (x- \alpha )|. sin 2 \alpha + C$