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Integrate the function $\large\int\frac{\sin x}{\sin(x-a)}$

1 Answer

  • (i) if given $I=\int f(x)dx,$ let f(x)=t, then $f'(x)dx=dt$,then
  • (ii)$\sin (A+B)=\sin A \cos B+\cos A\sin B$
  • (iii)$\cot x dx=\log|\sin x|+c$
Step 1:
Given $I=\large\int\frac{\sin x}{\sin(x-a)}$$dx$
Let $x-a=t,$ on integrating we get,
Therefore $x=t+a$
Substituting for t and dt we get,
$I=\large\int\frac{\sin (t+a)}{\sin t}$$dt$
But we know $\sin (A+B)=\sin A \cos B+\cos A \sin B$
Therefore $ I=\large\int \frac{\sin t \cos a+\cos t \sin a}{\sin t}$
Step 2:
On seperating the terms,
$I=\large\int \frac{\sin t \cos a}{\sin t}$$dt+\int\large \frac{\cos t \sin a}{\sin t}$$dt$
$\;\;\;=\int cos a\;dt+\int \sin a. \cot t\;dt$
$=\cos a\int dt+\sin a\int \cot t\;dt$
On integrating we get
$\cos a(t)+\sin a.\log|\sin t|+c$
Substituting for t we get
$\cos a(x-a)+\sin a.\log|\sin (x-a)|+c$
$=x\cos a+\sin a.\log|\sin (x-a)|+a \cos a+c$
Since $a \cos a$ is a constant
$=x\cos a+\sin a.\log|\sin (x-a)|+c$
answered Sep 10, 2013 by sreemathi.v