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