Question
Evaluate: $\int\frac{\sin(\text{x} - \text{a})}{\sin(\text{x + a})}\text{ dx}.$

Answer

Let $\text{I} =\int\frac{\sin(\text{x - a )}}{\sin\text{(x +a )}}\text{dx}$
Let x + a =t $\Rightarrow$x =t – a
$\Rightarrow$dx = dt
$\therefore\text{I} = \int\frac{\sin(\text{t} - 2\text{a})}{\sin\text{t}}\text{dt}$
$ =\int\frac{\sin\text{t}.\cos2\text{a} - \cos\text{t}.\sin2\text{a}}{\sin\text{t}}\text{dt}$
$= \cos 2\text{a} \int \text{dt} – \int \sin 2\text{a}. \cot \text{t dt} = \cos 2\text{a.t} – \sin 2\text{a}. \log|\sin \text{t}|+ \text{C}$
$= \cos 2\text{a}.\text{(x + a)} – \sin 2\text{a}. \log|\sin \text{(x + a)|+ C}$
$= \text{x} \cos 2\text{a + a} \cos 2\text{a} – (\sin 2\text{a)} \log|\sin \text{(x + a)|+ C}.$

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