$$\int\frac{\sin x}{\sin (x-a)\cdot \sin (x-b)}\,\mathrm dx$$

My Try::

\begin{align} &\displaystyle \frac{1}{\sin (b-a)}\int\frac{\sin \{(x-a)-(x-b)\}}{\sin (x-a)\cdot \sin (x-b)}\cdot\sin x\,\mathrm dx\\ =& \frac{1}{\sin (b-a)}\int \left\{\cot (x-b)-\cot (x-a)\right\}\cdot \sin x \,\mathrm dx \end{align}

Now how can I proceed after that?

  • $\begingroup$ Please do not use titles consisting only of math expressions; these are discouraged for technical reasons -- see meta. $\endgroup$ – Lord_Farin Sep 20 '13 at 19:27
  • $\begingroup$ Sorry Lord_Farin , next time i will take care of it. $\endgroup$ – juantheron Sep 20 '13 at 19:40
  • $\begingroup$ @experimentX Please do not bump questions with such minuscule edits of disputable quality. $\endgroup$ – Lord_Farin Sep 20 '13 at 20:32
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    $\begingroup$ You may start from $\sin(x)=\sin(x-a+a).$ $\endgroup$ – user64494 Sep 20 '13 at 20:33
  • $\begingroup$ Step 1: Open those $\sin(x-a)$ and $\sin(x-b)$ to write everything in terms of $\sin(x)$ and $\cos(x)$. $\endgroup$ – Pp.. Feb 7 '15 at 1:34

Wrong answer for now.

First let's rewrite: $$\frac{1}{\sin(x-a)\sin(x-b)}=\frac{p}{\sin(x-a)}+\frac{q}{\sin(x-b)} \quad(1)$$ Now multiplying $(1)$ by $\sin(x-a)$ followed by letting $x=a$ yields: $$\left(\frac{1}{\sin(x-b)}=p+q\underset{\rightarrow 0}{\frac{\sin(x-a)}{\sin(x-b)}}\right)\bigg|_{x=a}\Rightarrow p=\frac{1}{\sin(a-b)}$$ Similarly to obtain q, multiply $(1)$ by $\sin(x-b)$ and plugg $x=b$ gives: $$\left(\frac{1}{\sin(x-a)}=p\underset{\rightarrow 0}{\frac{\sin(x-b)}{\sin(x-a)}}+q\right)\bigg|_{x=b}\Rightarrow q=\frac{1}{\sin(b-a)}=-p$$ Back to the original integral, we have: $$I=\int\frac{\sin x}{\sin (x-a)\cdot \sin (x-b)}dx=\frac{1}{\sin(a-b)}\left(\int\frac{\sin x}{\sin (x-a)}dx-\int\frac{\sin x}{\sin(x-b)}dx\right)$$ Now, it's enough to solve only one of these integrals. $$\int\frac{\sin x}{\sin(x-t)}dx=\int\frac{\sin(x-t+t)}{\sin (x-t)}dx=\cos t\int\frac{\sin(x-t)}{\sin(x-t)}dx+\sin t\int\frac{\cos(x-t)}{\sin(x-t)}dx=$$ $$=\cos t \cdot x +\sin t \cdot \ln|\sin(x-t)|+c$$ Thus replacing $t$ by $a$ and $b$ we get the answer to be: $$I=\frac{1}{\sin(a-b)}\left((\cos a-\cos b)x+\sin a \ln|\sin(x-a)|-\sin b \ln|\sin(x-b)|\right)+C$$

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    $\begingroup$ Unfortunately,$$\frac{1}{\sin\left(a-b\right)}\left(\frac{1}{\sin\left(x-a\right)}-\frac{1}{\sin\left(x-b\right)}\right)=\frac{\sin\left(x-b\right)-\sin\left(x-a\right)}{\sin\left(a-b\right)\sin\left(x-a\right)\sin\left(x-b\right)}=\frac{\cos\left(x-\frac{a+b}{2}\right)}{\cos\frac{a-b}{2}\sin\left(x-a\right)\sin\left(x-b\right)}\not\equiv\frac{1}{\sin\left(x-a\right)\sin\left(x-b\right)}.$$ $\endgroup$ – J.G. Dec 20 '18 at 12:08
  • $\begingroup$ This is quite weird, I don't see where is my error. $\endgroup$ – Zacky Dec 20 '18 at 12:14
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    $\begingroup$ Your error is assuming constant coefficients exist that get the job done in the first place. $\endgroup$ – J.G. Dec 20 '18 at 12:48

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