Question

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

• 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 1

The correct option is A $x-\alpha$ and ${\log }_e|\sin (x-\alpha )|$

$\int \frac{\tan x+\tan \alpha }{\tan x-\tan \alpha }dx$
$\Rightarrow \int \frac{\sin x\cos \alpha +\cos x\sin \alpha }{\sin x\cos \alpha -\cos x\sin \alpha }dx=\int \frac{\sin (x+\alpha )}{\sin (x-\alpha )}dx=\int \frac{\sin (x-\alpha +2\alpha )}{\sin (x-\alpha )}dx=\int \frac{\sin (x-\alpha )\cos 2\alpha }{\sin (x-\alpha )}dx+\int \frac{\cos (x-\alpha )\sin 2\alpha }{\sin (x-\alpha )}dx=x\cos 2\alpha +\sin 2\alpha \log |\sin (x-\alpha )|+c$

Here, option $\left(a\right)$ and $\left(b\right)$ both are correct so it is bonus.