I'm stuck in solving the integral of $\dfrac{1}{\sin(x-a)\sin(x-b)}$. I "developed" the sin at denominator and then I divided it by $\cos^2x$ obtaining $$\int\frac{1}{\cos(a)\cos(b)\operatorname{tan}^2x-\cos(a)\sin(b)\operatorname{tan}x-\sin(a)\cos(b)\operatorname{tan}x+\sin(a)\sin(b)}\frac{1}{\cos^2x}dx$$ Then I made a substitution by $t=\operatorname{tan}x$ arriving to this $$\int\frac{1}{\cos(a)\cos(b)t^2-(\cos(a)\sin(b)+\sin(a)\cos(b))t+\sin(a)\sin(b)}dt$$ How can I solve it now? (probably I forgot something, it easy to make mistakes here)

Thank you in advance!

  • $\begingroup$ $\frac{d}{du}\cot u=-\csc^2 u$ $\endgroup$
    – Jonathan
    May 2 '14 at 15:17
  • $\begingroup$ Do you really mean to have $\sin(x-a) \sin(x-a)$ in the denominator? If so you're just integrating csc^2. $\endgroup$
    – coffeemath
    May 2 '14 at 15:17
  • $\begingroup$ Sorry I wrote it wrong!! $\endgroup$
    – Dipok
    May 2 '14 at 15:21
  • $\begingroup$ I edited your question. Format stuff. See if everything is all right. And use \sin, \cos instead sin, cos. For tg I used \operatorname{tg} because there is no $\LaTeX$ command for that (on other side, there is \tan for $\tan$, which is same, but there is difference from country to country (in my country they also use tg, for example)) $\endgroup$
    – Cortizol
    May 2 '14 at 15:37

Here is another approach to join the 'party' using a simple trigonometric technique. $$ \begin{align} &\int \frac{1}{\sin(x-a)\sin(x-b)} dx\\&=\frac{1}{\sin(a-b)}\int \frac{\sin(a-b)}{\sin(x-a)\sin(x-b)} dx\\ &=\frac{1}{\sin(a-b)}\int \frac{\sin((x-b)-(x-a))}{\sin(x-a)\sin(x-b)} dx\\ &=\csc(a-b)\int \frac{\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a)}{\sin(x-a)\sin(x-b)} dx\\ &=\csc(a-b)\left[\int \frac{\sin(x-b)\cos(x-a)}{\sin(x-a)\sin(x-b)} dx-\int \frac{\cos(x-b)\sin(x-a)}{\sin(x-a)\sin(x-b)} dx\right]\\ &=\csc(a-b)\left[\int \frac{\cos(x-a)}{\sin(x-a)} dx-\int \frac{\cos(x-b)}{\sin(x-b)} dx\right]\\ &=\csc(a-b)\left[\int \frac{1}{\sin(x-a)}d(\sin(x-a))-\int \frac{1}{\sin(x-b)} d(\sin(x-b))\right]\\ &=\csc(a-b)\bigg[\ln|\sin(x-a)|-\ln|\sin(x-b)|\bigg]+\text{C}\\ &=\csc(a-b)\ln\left|\frac{\sin(x-a)}{\sin(x-b)}\right|+\text{C}.\\ \end{align} $$

  • 2
    $\begingroup$ Very nice! (+1) $\endgroup$
    – Hakim
    May 2 '14 at 16:26
  • 1
    $\begingroup$ +1 Mr. Tunk-Fey! You always have a different answer yet very elegant. b(^_^)d $\endgroup$ May 2 '14 at 16:52
  • 2
    $\begingroup$ Thanks all. Special thanks to @V-Moy for the editing. :) $\endgroup$
    – Tunk-Fey
    May 2 '14 at 16:54

Here is another way to perform the integral if you don't mind using complex numbers.

Let $z = e^{ix}, \alpha = e^{ia}$ and $\beta = e^{ib}$, we have

$$\begin{align} & \int \frac{1}{\sin(x-a)\sin(x-b)} dx\\ = & \int \frac{1}{ \left(\frac{z\alpha^{-1}-\alpha z^{-1}}{2i}\right) \left(\frac{z\beta^{-1}-\beta z^{-1}}{2i}\right) }\frac{dz}{zi}\\ = & 4i\alpha\beta\int\frac{z dz}{(z^2-\alpha^2)(z^2-\beta^2)}\\ = & \frac{4i\alpha\beta}{\alpha^2-\beta^2} \int\left(\frac{z}{z^2-\alpha^2} -\frac{z}{z^2-\beta^2}\right) dz\\ = & \frac{2i}{\alpha\beta^{-1} - \beta\alpha^{-1}}\left( \log(z^2 - \alpha^2) - \log(z^2-\beta^2)\right) + \text{const}\\ = & \frac{1}{\sin(a-b)} \log\left(\frac{z\alpha^{-1} - \alpha z^{-1}}{z\beta^{-1} - \beta z^{-1}}\right) + \text{const}'\\ =& \frac{1}{\sin(a-b)}\log\frac{\sin(x-\alpha)}{\sin(x-\beta)} + \text{const}' \end{align} $$


Hint for the original problem $\int\frac{1}{\sin(x-a)\sin(x-a)} \mathrm{d}x$:

$$\frac{1}{\sin(x-a)\sin(x-a)}=\csc^2(x-a) \; \mathrm{and} \, \int{\csc^2u \,\mathrm{d}u}=-\cot u+C$$

  • $\begingroup$ I clarified that this was for the original problem otherwise it may look weird. $\endgroup$
    – wythagoras
    Oct 27 '15 at 13:34

Looks like I am late in the race. Let me present you a slight different way to integrate it.

let $\displaystyle x = y + \frac{a+b}{2}$, and also let $\displaystyle \frac{a-b}{2} = c$ $$\int \frac{1}{ \sin\left(y - \frac{a-b}{2}\right)\sin\left(y + \frac{a-b}{2}\right)} \, dx = \int \frac{1}{\sin(y-c) \sin(y+c)} \, dy $$

Expanding this we get, $$\int \frac{1}{\sin^2(y) \cos^2(c) - \cos^2(y)\sin^2(c)}dy = \int \frac{\csc^2 (c) \sec^2(y) }{\tan^2(y) \cot^2(c) - 1} \, dy $$

Let, $\cot(c)\tan(y) = z$, then we get, $$\frac{1}{2 \cos(c)\sin(c)} \int \frac{1}{z^2 - 1}dz=\frac{1}{2 \cos(c)\sin(c)} \log \left( \frac{z-1}{z+1} \right ) $$

Substituting values, we get $$\frac{1}{\sin(a-b)} \log \left( \frac{ \sin \left( -\frac{a-b}{2} + \left(x- \frac{a+b}{2}\right ) \right )}{ \sin \left( \frac{a-b}{2} + \left( x- \frac{a+b}{2}\right ) \right )} \right ) = \color{red}{\frac{1}{\sin(a-b)} \log \left( \frac{\sin(x-a)}{\sin(x-b)} \right ) + \mathrm{constant}}$$


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