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Let $\alpha \epsilon (0,\pi /2)$ be fixed. If the integral $\int \frac{\tan x+\tan \alpha }{\tan x-\tan \alpha}dx=$ 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 :

Option 1)
$x+\alpha$ and $\log_{e}|\sin (x+\alpha )|$

Option 2)
$x-\alpha$ and $\log_{e}|\sin (x-\alpha )|$

Option 3)
$x-\alpha$ and $\log_{e}|\cos (x-\alpha )|$

Option 4)
$x+\alpha$ and $\log_{e}|\sin (x-\alpha )|$

Answers (1)

I = $\int \frac{\tan x+\tan \alpha }{\tan x-\tan \alpha}dx=$ $A(x)\cos 2\alpha +B(x)\sin 2\alpha +C$

$I= \int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha }{\cos \alpha }}{\frac{\sin x}{\cos x}-\frac{\sin \alpha }{\cos \alpha }}dx$

$=\int \frac{\sin x\cos \alpha +\sin \alpha \cos x}{\sin x\cos \alpha -\sin \alpha \cos x}dx$

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

$=\int \frac{\sin (x+\alpha )}{\sin[(x+\alpha )-2\alpha ]}dx$

$=\int \frac{\sin (x+\alpha) }{\sin(x+\alpha )\sin {2}\alpha -\sin (x-\alpha \alpha )\sin {2}\alpha }dx$

$=(x-\alpha )\cos{2}\alpha +\log_{e}|\sin (x-\alpha )|\sin {2}\alpha +C$

comparing with LHS $\Rightarrow$

$\left [ \begin{matrix} A(x)=x-\alpha \\ B(x)=\log_{e}|\sin (x-\alpha )| \end{matrix} \right ]$

Option 1)

$x+\alpha$ and $\log_{e}|\sin (x+\alpha )|$

Option 2)

$x-\alpha$ and $\log_{e}|\sin (x-\alpha )|$

Option 3)

$x-\alpha$ and $\log_{e}|\cos (x-\alpha )|$

Option 4)

$x+\alpha$ and $\log_{e}|\sin (x-\alpha )|$

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Vakul

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