Gujarat 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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