Let alpha in (0, pi /2) be fixed. If the inte | vedclass

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(C) We are given the integral $I = \int \frac{	an x + 	an \alpha}{	an x - 	an \alpha} dx$.Using the identity $	an x \pm 	an \alpha = \frac{\sin(

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$x - \alpha$ and $\log_e |\sin (x - \alpha)|$

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(C) We are given the integral $I = \int \frac{	an x + 	an \alpha}{	an x - 	an \alpha} dx$.Using the identity $	an x \pm 	an \alpha = \frac{\sin(x \pm \alpha)}{\cos x \cos \alpha}$,we have:$I = \int \frac{\sin(x + \alpha) / (\cos x \cos \alpha)}{\sin(x - \alpha) / (\cos x \cos \alpha)} dx = \int \frac{\sin(x + \alpha)}{\sin(x - \alpha)} dx$.Let $t = x - \alpha$,then $x = t + \alpha$ and $dx = dt$. Substituting these into the integral:$I = \int \frac{\sin(t + 2\alpha)}{\sin t} dt$.Using the expansion $\sin(t + 2\alpha) = \sin t \cos 2\alpha + \cos t \sin 2\alpha$:$I = \int \frac{\sin t \cos 2\alpha + \cos t \sin 2\alpha}{\sin t} dt$$I = \int \cos 2\alpha dt + \int \cot t \sin 2\alpha dt$$I = t \cos 2\alpha + \sin 2\alpha \ln |\sin t| + C$.Substituting $t = x - \alpha$ back:$I = (x - \alpha) \cos 2\alpha + \ln |\sin (x - \alpha)| \sin 2\alpha + C$.Comparing this with $A(x) \cos 2\alpha + B(x) \sin 2\alpha + C$,we get $A(x) = x - \alpha$ and $B(x) = \ln |\sin (x - \alpha)|$.Thus,the correct option is $(C)$.

Let $\alpha \in (0, \pi /2)$ 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:

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