Question
Evalute the following integrals:
$\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$

Answer

Let $\text{I}=\int\frac{\sin(\text{x}-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$ then,
$\text{I}=\int\frac{\sin(\text{x}-\alpha+\alpha-\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha-2\alpha)}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\frac{\sin(\text{x}+\alpha)\cos2\alpha-\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\text{dx}$
$=\int\Big[\frac{\sin(\text{x}+\alpha)\cos2\alpha}{\sin(\text{x}+\alpha)}-\frac{\cos(\text{x}+\alpha)\sin2\alpha}{\sin(\text{x}+\alpha)}\Big]\text{dx}$
$=\int\big(\cos2\alpha-\cot(\text{x}+\alpha)\sin2\alpha\big)\text{dx}$
$=\cos2\alpha\int\text{dx}-\sin2\alpha\int\cot(\text{x}+\alpha)\text{dx}$
$=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$
$\therefore\text{I}=\text{x}\cos2\alpha-\sin2\alpha\log|\sin(\text{x}+\alpha)|+\text{C}$

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