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

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, 2013 at 19:27
  • $\begingroup$ Sorry Lord_Farin , next time i will take care of it. $\endgroup$
    – juantheron
    Sep 20, 2013 at 19:40
  • $\begingroup$ @experimentX Please do not bump questions with such minuscule edits of disputable quality. $\endgroup$
    – Lord_Farin
    Sep 20, 2013 at 20:32
  • 1
    $\begingroup$ You may start from $\sin(x)=\sin(x-a+a).$ $\endgroup$
    – user64494
    Sep 20, 2013 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, 2015 at 1:34

1 Answer 1


$$I_{a,b}(x)=\frac{1}{\sin(b-a)}\int\frac{\sin \left((x-a)-(x-b)\right)}{\sin (x-a)\cdot \sin (x-b)} \sin(x)\ dx$$ Angle-addition: $$=\frac{1}{\sin (b-a)}\int \left(\cot (x-b)-\cot (x-a)\right)\cdot \sin (x) \, dx$$ $$=\frac{1}{\sin (b-a)}\left(\int \cot (x-b) \sin x \, dx -\int \cot (x-a) \sin (x) \, dx\right)$$ For the first integral, let $x=z+b$: $$\int \cot(z)\sin(z+b)\,dz$$ Angle-addition: $$=\int\cos(z)\cos(b)+\cos^2(z)/\sin(z)\sin(b)\,dz$$ $$=\sin(z)\cos(b)+\sin(b)\left(\cos(z)+\log\left(\sin\left(\frac{z}{2}\right)\right)-\log\left(\cos\left(\frac{z}{2}\right)\right)\right)C$$ $$\Rightarrow \sin(x-b)\cos(b)+\sin(b)\left(\cos(b-x)+\log\left(\sin\left(\frac{x-b}{2}\right)\right)-\log\left(\cos\left(\frac{x-b}{2}\right)\right)\right)+C$$ The other side is similar.


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