Find the integral: $\int \frac{\sin x}{\sin(x + a)} \, dx$

$x \cos a - \sin a \log |\sin(x + a)| + C$

The correct answer is Option (4) → $x \cos a - \sin a \log |\sin(x + a)| + C$

Put $x + a = t$. Then $dx = dt$. Therefore

$\int \frac{\sin x}{\sin(x + a)} \, dx = \int \frac{\sin(t - a)}{\sin t} \, dt$

$= \int \frac{\sin t \cos a - \cos t \sin a}{\sin t} \, dt$

$= \cos a \int dt - \sin a \int \cot t \, dt$

$= (\cos a) t - (\sin a) \left[ \log |\sin t| + C_1 \right]$

$= (\cos a) (x + a) - (\sin a) \left[ \log |\sin(x + a)| + C_1 \right]$

$= x \cos a + a \cos a - (\sin a) \log |\sin(x + a)| - C_1 \sin a$

Hence, $\int \frac{\sin x}{\sin(x + a)} \, dx = x \cos a - \sin a \log |\sin(x + a)| + C,$

where, $C = -C_1 \sin a + a \cos a,$ is another arbitrary constant.