Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSIntegrals1 Mark
Question
Find the integral: $\int \frac{\sin x}{\sin (x+a)} d x$
✓
Answer
Put $x + a = t.$ Then $dx = dt.$ Therefore
$\int \frac{\sin x}{\sin (x+a)} d x=\int \frac{\sin (t-a)}{\sin t} d t$
$= \int \frac{\sin t \cos a-\cos t \sin a}{\sin t} d t$
$= \cos a \int d t-\sin a \int \cot t d t$
$= (\cos a) t-(\sin a)\left[\log |\sin t|+C_{1}\right]$
$= (\cos a)(x+a)-(\sin a)\left[\log |\sin (x+a)|+C_{1}\right]$
$= x \cos a + a \cos a - (\sin a) \log |\sin (x + a)| - C_1 \sin a$
Hence, $\int \frac{\sin x}{\sin (x+a)} d x = x \cos a - (\sin a) \log |\sin (x + a)| - C, $
where, $C = -C_1 \sin a + a \cos a,$ is another arbitrary constant.
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