91Ó°ÊÓ

Using the hint given in the exercise, we can rewrite \(\sin 2y\) as: $$\sin 2y = 2\sin y \cos y$$ Now, we can replace this in the integral expression: $$\int_{0}^{\pi / 6} \frac{2\sin y \cos y}{\sin ^{2} y+2} d y$$

Now, let's divide both numerator and denominator by \(\cos^2 y\): $$\int_{0}^{\pi / 6} \frac{2\sin y \cos y}{\sin ^{2} y+2} d y = \int_{0}^{\pi / 6} \frac{\frac{2\sin y \cos y}{\cos^2 y}}{\frac{\sin ^{2} y+2}{\cos^2 y}} d y$$ We get: $$\int_{0}^{\pi / 6} \frac{2\sin y \cos y}{\sin ^{2} y+2} d y = \int_{0}^{\pi / 6} \frac{2\tan y}{\tan ^{2} y+2} d y$$

Next, we apply the substitution technique: Let \(u = \tan y\). Then, we have \(\frac{du}{dy} = \sec^2 y = 1 + \tan^2 y = 1 + u^2\). So, \(dy = \frac{du}{1+u^2}\). Also, note that when \(y=0\), \(u=\tan 0=0\), and when \(y=\frac{\pi}{6}\), \(u=\tan \frac{\pi}{6} = \sqrt{3}/3\). Now, we can rewrite the integral as: $$\int_{0}^{\frac{\pi}{6}} \frac{2\tan y}{\tan ^{2} y+2} d y = \int_{0}^{\sqrt{3}/3} \frac{2u}{u ^{2}+2} \frac{du}{1+u^2}$$

Now, we simplify the integral and evaluate it: $$\int_{0}^{\sqrt{3}/3} \frac{2u}{(1+u^2)(u^{2}+2)} du$$ Let's split the integral into two simpler integrals: $$\int_{0}^{\sqrt{3}/3} \left(\frac{A}{1+u^2} + \frac{B}{u^2+2}\right) du$$ Setting \(u=0\), we get \(A=1\). Next, we set \(u^2=-1\), and we get \(B=-1\). So, we have: $$\int_{0}^{\sqrt{3}/3} \left(\frac{1}{1+u^2} - \frac{1}{u^2+2}\right) du$$ Now, we can evaluate the two integrals separately: $$\left[\arctan u - \frac{1}{\sqrt{2}} \arctan\frac{u}{\sqrt{2}}\right]_{0}^{\sqrt{3}/3}$$ Finally, we substitute the values and calculate the result: $$\left[\arctan\frac{\sqrt{3}}{3} - \frac{1}{\sqrt{2}} \arctan\frac{\sqrt{3}/3}{\sqrt{2}}\right] - \left[\arctan 0 - \frac{1}{\sqrt{2}} \arctan 0\right]$$ $$=\arctan\frac{\sqrt{3}}{3} - \frac{1}{\sqrt{2}} \arctan\frac{\sqrt{3}}{3}$$