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

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

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$	ext { Let } \begin{aligned} \mathrm{I} & =\int \frac{	an x+	an \alpha}{	an x-	an \alpha} \mathrm{d} x \ & =\int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x}-\frac{\sin \alpha}{\cos \alpha}} \mathrm{d} x \ & =\int \frac{\sin x \cos \alpha+\sin \alpha \cos x}{\sin x \cos \alpha-\sin \alpha \cos x} \mathrm{~d} x \ & =\int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} \mathrm{d} x \end{aligned}$ Let $x-\alpha=\mathrm{t}$ $\begin{aligned} 	herefore \quad \mathrm{I} & =\frac{\sin (\mathrm{t}+2 \alpha)}{\sin \mathrm{t}} \ & =\int \frac{\sin (\mathrm{t}) \cos 2 \alpha+\cos (\mathrm{t}) \sin 2 \alpha}{\sin (\mathrm{t})} \mathrm{d} x \ & =\cos 2 \alpha \int 1 \mathrm{dt}+\sin 2 \alpha \int \cot (\mathrm{t}) \mathrm{dt} \ & =\cos 2 \alpha \cdot \mathrm{t}+\sin 2 \alpha \cdot \log |\sin (\mathrm{t})|+\mathrm{c} \ 	herefore \quad \mathrm{I} & =(x-\alpha) \cos 2 \alpha+\log |\sin (x-\alpha)| \sin 2 \alpha+\mathrm{c} \ & \mathrm{But} \int \frac{	an x+	an \alpha}{	an x-	an \alpha} \mathrm{d} x=\mathrm{A}(x) \cos 2 \alpha+\mathrm{B}(x) \sin 2 \alpha+\mathrm{c} ... [Given] \ & \Rightarrow \mathrm{A}(x)=x-\alpha, \mathrm{B}(x)=\log |\sin (x-\alpha)|+\mathrm{c} \end{aligned}$

Let $\alpha \in\left(0, \frac{\pi}{2}\right)$ be fixed. If the integral $\int \frac{\tan x+\tan \alpha}{\tan…

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