# then the functions A(x) and B(x) are respectively:

Question:

Let $\alpha \in(0, \pi / 2)$ be fixed. If the integral

$\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} d x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\mathrm{C}$, where $\mathrm{C}$ is

a constant of integration, then the functions $\mathrm{A}(x)$ and $\mathrm{B}(x)$ are respectively:

1. (1) $x+\alpha$ and $\log _{e}|\sin (x+\alpha)|$

2. (2) $x-\alpha$ and $\log _{e}|\sin (x-\alpha)|$

3. (3) $x-\alpha$ and $\log _{e}|\cos (x-\alpha)|$

4. (4) $x+\alpha$ and $\log _{e}|\sin (x-\alpha)|$

Correct Option: , 2

Solution:

Given integral, $I=\int \frac{\left(2 x^{3}-1\right) d x}{x^{4}+x}=\int \frac{\left(2 x-x^{-2}\right) d x}{x^{2}+x^{-1}}$

$\int \frac{\tan x+\tan \alpha}{\tan x-\tan \alpha} d x=\int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} d x$

Let $x-\alpha=t \Rightarrow d x=d t$

$=\int \frac{\sin (t+2 \alpha)}{\sin t} d t=\int[\cos 2 \alpha+\sin 2 \alpha \cdot \cot t] d t$

$=\cos 2 \alpha \cdot t+\sin 2 \alpha \cdot \log |\sin t|+c$

$=(x-\alpha) \cdot \cos 2 \alpha+\sin 2 \alpha \cdot \log |\sin (x-\alpha)|+c$