Let alpha in (0, frac{pi}{2}) be fixed. If th | vedclass

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(B) Let $I = \int \frac{	an x + 	an \alpha}{	an x - 	an \alpha} dx$. $= \int \frac{\frac{\sin x}{\cos x} + \frac{\sin \alpha}{\cos \alpha}}{\frac{

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$x-\alpha$ and $\log |\sin (x-\alpha)|$.

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(B) Let $I = \int \frac{	an x + 	an \alpha}{	an x - 	an \alpha} dx$. $= \int \frac{\frac{\sin x}{\cos x} + \frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x} - \frac{\sin \alpha}{\cos \alpha}} dx$ $= \int \frac{\sin x \cos \alpha + \sin \alpha \cos x}{\sin x \cos \alpha - \sin \alpha \cos x} dx$ $= \int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} dx$. Let $t = x - \alpha$,so $x = t + \alpha$ and $dx = dt$. $I = \int \frac{\sin (t + 2\alpha)}{\sin t} dt$ $= \int \frac{\sin t \cos 2\alpha + \cos t \sin 2\alpha}{\sin t} dt$ $= \cos 2\alpha \int 1 dt + \sin 2\alpha \int \cot t dt$ $= t \cos 2\alpha + \sin 2\alpha \log |\sin t| + c$ $= (x - \alpha) \cos 2\alpha + \log |\sin (x - \alpha)| \sin 2\alpha + c$. Comparing this with $A(x) \cos 2\alpha + B(x) \sin 2\alpha + c$,we get $A(x) = x - \alpha$ and $B(x) = \log |\sin (x - \alpha)|$.

Let $\alpha \in (0, \frac{\pi}{2})$ 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 functions $A(x)$ and $B(x)$ are respectively

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