Integrating Fractional Sines

$\begin{aligned} I & = \int_{-\frac \pi 6}^\frac \pi 6 \frac {\sin 2x}{\sin 3x} dx & \small \color{#3D99F6} \frac {\sin 2x}{\sin 3x} \text{ is an even function.} \\ & = 2 \int_0^\frac \pi 6 \frac {\sin 2x}{\sin 3x} dx \\ & = 2 \int_0^\frac \pi 6 \frac {2\sin x\cos x}{3\sin x - 4\sin^3 x} dx & \small \color{#3D99F6} \text{Let }u = \sin x \implies du = \cos x\ dx \\ & = 4 \int_0^\frac 12 \frac u{3u - 4u^3} du \\ & = \frac 43 \int_0^\frac 12 \frac 1{1 - \frac 43 u^2} du & \small \color{#3D99F6} \text{Let }\sin \theta = \frac 2{\sqrt 3}u \implies \cos \theta\ d\theta = \frac 2{\sqrt 3}\ du \\ & = \frac 2{\sqrt 3} \int_0^{\sin^{-1}\frac 1{\sqrt 3}} \sec \theta \ d\theta \\ & = \frac 2{\sqrt 3} \ln(\tan \theta + \sec \theta) \bigg|_0^{\sin^{-1}\frac 1{\sqrt 3}} \\ & = \frac 2{\sqrt 3} \left(\ln\left(\frac 1{\sqrt 2} + \sqrt{\frac 32}\right)- \ln (0+1) \right) \\ & = \frac 1{\sqrt 3}\ln \left(\frac {1+\sqrt 3}{\sqrt 2}\right)^2 \\ & = \frac {\ln \left(2+\sqrt 3\right)}{\sqrt 3} \\ & \approx \boxed{0.760} \end{aligned}$

$\displaystyle\int_{-\cfrac{\pi}{6}}^{\cfrac{\pi}{6}} \cfrac{sin2x}{sin3x} dx$

Let $f(x)=\cfrac{sin2x}{sin3x}$

$f(-x)=\cfrac{sin(-2x)}{sin(-3x)}=\cfrac{-sin2x}{-sin3x}=\cfrac{sin2x}{sin3x}=f(x)$

$f(x)=\cfrac{sin2x}{sin3x}$ is even function.

$\therefore \displaystyle\int_{-\cfrac{\pi}{6}}^{\cfrac{\pi}{6}} \cfrac{sin2x}{sin3x}dx=2\displaystyle\int_{0}^{\cfrac{\pi}{6}} \cfrac{sin2x}{sin3x}dx=2\displaystyle\int_{0}^{\cfrac{\pi}{6}} \cfrac{2sinxcosx}{3sinx-4sin^3x}dx=2\displaystyle\int_{0}^{\cfrac{\pi}{6}} \cfrac{2cosx}{3-4sin^2x}dx=4\displaystyle\int_{0}^{\cfrac{\pi}{6}} \cfrac{cosx}{3-4sin^2x}dx$

Let $y=sinx\Rightarrow dy=d(sinx)=cosxdx$

When $x=\cfrac{\pi}{6}, y=sin\cfrac{\pi}{6}=\cfrac{1}{2}$

When $x=0, y=sin0=0$

$\displaystyle\int_{-\cfrac{\pi}{6}}^{\cfrac{\pi}{6}} \cfrac{sin2x}{sin3x}dx=4\displaystyle\int_{0}^{\cfrac{1}{2}} \cfrac{1}{3-4y^2}dy=4\displaystyle\int_{0}^{\cfrac{1}{2}} \cfrac{1}{(\sqrt{3}+2y)(\sqrt{3}-2y)}dy=\cfrac{4}{2\sqrt{3}}\displaystyle\int_{0}^{\cfrac{1}{2}} (\cfrac{1}{\sqrt{3}+2y}+\cfrac{1}{\sqrt{3}-2y})dy$

$=\cfrac{2}{\sqrt{3}}[\cfrac{ln(\sqrt{3}+2y)}{2}+\cfrac{ln(\sqrt{3}-2y)}{-2}]_{0}^{\cfrac{1}{2}}=\cfrac{1}{\sqrt{3}}[ln(\sqrt{3}+2y)-ln(\sqrt{3}-2y)]_{0}^{\cfrac{1}{2}}=\cfrac{1}{\sqrt{3}}[ln\cfrac{\sqrt{3}+2y}{\sqrt{3}-2y}]_{0}^{\cfrac{1}{2}}$

$=\cfrac{1}{\sqrt{3}}(ln\cfrac{\sqrt{3}+1}{\sqrt{3}-1}-ln\cfrac{\sqrt{3}}{\sqrt{3}})=\cfrac{1}{\sqrt{3}}(ln\cfrac{\sqrt{3}+1}{\sqrt{3}-1}-ln1)=\cfrac{1}{\sqrt{3}}(ln\cfrac{\sqrt{3}+1}{\sqrt{3}-1}-0)\approx0.76035$