Single Choice

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

A$$x - \alpha$$ and $$\log_{e} |\cos (x - \alpha)|$$
B$$x + \alpha$$ and $$\log_{e} |\sin (x - \alpha)|$$
C$$x - \alpha$$ and $$\log_{e} |\sin (x - \alpha)|$$
Correct Answer
D$$x + \alpha$$ and $$\log_{e} |\sin (x + \alpha)|$$

Solution

$$\int \dfrac {\tan x + \tan \alpha}{\tan x - \tan \alpha} dx = \int \dfrac {\sin (x + \alpha)}{\sin (x - \alpha)}dx$$
Let $$x - \alpha = t$$
$$\Rightarrow \int \dfrac {\sin (t + 2\alpha)}{\sin t}dt = \int \cos 2\alpha dt + \int \cot (t)\sin 2\alpha dt$$
$$= t.\cos 2\alpha + ln |\sin t|.\sin 2\alpha + C$$
$$= (x - \alpha)\cos 2\alpha + ln|\sin (x - \alpha)|.\sin 2\alpha + C$$.

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