The Question$$\int \csc\left(x-\frac{\pi}{3}\right)\csc\left(x-\frac{\pi}{6}\right) dx $$What I Tried-
I tried dividing both numerator and denominator by $\sin \pi/6$ but couldn't get too far, what I got after dividing and doing some simplification-$$2\int \frac{1}{\frac{\sqrt3}{2} - \sin 2x}dx \ = \int \csc\left(\frac{\pi}{6}+x\right)\sec\left(\frac{\pi}{6}-x\right)dx$$Now I am stuck after this step and don't think there is scope for simplification. Kindly tell where I am going wrong, it may be a very stupid mistake.
3 Answers
$\begingroup$Hint
$$\csc(x-\pi/3)\csc(x-\pi/6)$$
$$=\dfrac1{\sin(\pi/3-\pi/6)}\cdot\dfrac{\sin(x-\pi/6-(x-\pi/3))}{\sin(x-\pi/3)\sin(x-\pi/6)}$$
$\endgroup$ 1 $\begingroup$$$\csc \left(x-\frac{\pi }{3}\right)\csc \left(x-\frac{\pi }{6}\right)= \csc \left(\frac{\pi }{6}-x\right) \sec \left(x+\frac{\pi }{6}\right)=\frac{4}{\sqrt{3}-2 \sin (2x) }$$ Now, let $x=\tan^{-1}(t)$ and tou face$$I=4\int \frac{dt}{\sqrt{3} t^2-4 t+\sqrt{3}}$$ The denominator has two simple real roots. Then partial fraction decomposition to face two simple integrals.
$\endgroup$ 2 $\begingroup$Substitute $t= \tan(x-\frac\pi4)$
$$\int \csc\left(x-\frac{\pi}{3}\right)\csc\left(x-\frac{\pi}{6}\right) dx =4\int \frac{2-\sqrt3}{t^2-(2-\sqrt3)^2}dt\\ = 4\ln | \frac{2-\sqrt3-t}{2-\sqrt3+ t} | = 4\ln | \frac{2-\sqrt3-\tan(x-\frac\pi4)}{2-\sqrt3+ \tan(x-\frac\pi4)} |+C $$
$\endgroup$
