Let $\alpha \in(0, \pi / 2)$ be fixed. If the integral $\int \frac{ an x+ an \alpha}{ an x- an \

Question

Let $\alpha \in(0, \pi / 2)$ 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 $C$ is constant of integration, then the functions $A(x)$ and $B(x)$ are respectively

Professor PC Thomas Classes · Answer

(b, c) : Let $I=\int \frac{	an x+	an \alpha}{	an x-	an \alpha} d x$\ $=\int \frac{\frac{\sin x}{\cos x}+\frac{\sin \alpha}{\cos \alpha}}{\frac{\sin x}{\cos x}-\frac{\sin \alpha}{\cos \alpha}} d x$\ $=\int \frac{\sin x \cos \alpha+\sin \alpha \cos x}{\sin x \cos \alpha-\sin \alpha \cos x} d x$\ $=\int \frac{\sin (x+\alpha)}{\sin (x-\alpha)} d x$\ Put $x-\alpha=z \Rightarrow d x=d z$\ $	herefore I=\int \frac{\sin (z+2 \alpha)}{\sin z} d z$\ $=\int \frac{\sin z \cos 2 \alpha+\cos z \sin 2 \alpha}{\sin z} d z$\ $=\int \cos 2 \alpha d z+\int \cot z \sin 2 \alpha d z$\ $=z \cos 2 \alpha+\sin 2 \alpha \log _{e}|\sin z|+C$\ $=(x-\alpha) \cos 2 \alpha+\sin 2 \alpha \log _{e}|\sin (x-\alpha)|+C$\ $	herefore A(x)=(x-\alpha)$ and $B(x)=\log _{e}|\sin (x-\alpha)|$\ If we consider $C=C+2 \alpha \cos 2 \alpha$, then $A(x)$ $=x+\alpha$\