# Unable to evaluate the integral $\int \csc\left(x-\frac{\pi}{3}\right)\csc\left(x-\frac{\pi}{6}\right) dx$

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.

• Are you sure it's not $\csc (x + \pi/6) \csc (x - \pi/6)$? Mar 27, 2021 at 4:39
• You are asking it for the question or for my answer? Mar 27, 2021 at 4:40
• I have multiplied the integral by 2, it is one and the same thing. Mar 27, 2021 at 4:46

Hint

$$\csc(x-\pi/3)\csc(x-\pi/6)$$

$$=\dfrac1{\sin(\pi/3-\pi/6)}\cdot\dfrac{\sin(x-\pi/6-(x-\pi/3))}{\sin(x-\pi/3)\sin(x-\pi/6)}$$

• I just figured that out, thanks! Mar 27, 2021 at 4:45

$$\csc \left(x-\frac{\pi }{3}\right)\csc \left(x-\frac{\pi }{6}\right)= \csc \left(\frac{\pi }{6}-x\right) \sec \left(x+\frac{\pi }{6}\right)=\frac{4}{\sqrt{3}-2 \sin (2x) }$$ 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.

• How did you think that, that's nowhere near standard substitutions Mar 27, 2021 at 4:55
• @VyomYadav. This is just the tangent half-angle subsitution. Mar 27, 2021 at 4:57

Substitute $$t= \tan(x-\frac\pi4)$$

$$\int \csc\left(x-\frac{\pi}{3}\right)\csc\left(x-\frac{\pi}{6}\right) dx =4\int \frac{2-\sqrt3}{t^2-(2-\sqrt3)^2}dt\\ = 4\ln | \frac{2-\sqrt3-t}{2-\sqrt3+ t} | = 4\ln | \frac{2-\sqrt3-\tan(x-\frac\pi4)}{2-\sqrt3+ \tan(x-\frac\pi4)} |+C$$