Lorraine Harvey

2021-12-31

The Question
$\int \mathrm{csc}\left(x-\frac{\pi }{3}\right)\mathrm{csc}\left(x-\frac{\pi }{6}\right)dx$
What I Tried- I tried dividing both numerator and denominator by $\mathrm{sin}\frac{\pi }{6}$ but couldnt

boronganfh

Hint
$\mathrm{csc}\left(x-\frac{\pi }{3}\right)\mathrm{csc}\left(x-\frac{\pi }{6}\right)$
$=\frac{1}{\mathrm{sin}\left(\frac{\pi }{3}-\frac{\pi }{6}\right)}\cdot \frac{\mathrm{sin}\left(x-\frac{\pi }{6}-\left(x-\frac{\pi }{3}\right)\right)}{\mathrm{sin}\left(x-\frac{\pi }{3}\right)\mathrm{sin}\left(x-\frac{\pi }{6}\right)}$

hysgubwyri3

$\mathrm{csc}\left(x-\frac{\pi }{3}\right)\mathrm{csc}\left(x-\frac{\pi }{6}\right)=\mathrm{csc}\left(\frac{\pi }{6}-x\right)\mathrm{sec}\left(x+\frac{\pi }{6}\right)=\frac{4}{\sqrt{3}-2\mathrm{sin}\left(2x\right)}$
Now, let $x={\mathrm{tan}}^{-1}\left(t\right)$ and tou face
$I=4\int \frac{dt}{\sqrt{3}{t}^{2}-4t+\sqrt{3}}$
The denominator has two simple real roots. Then partial fraction decomposition to face two simple integrals.

Vasquez

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

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