Select all expressions matching the value of the definite integral int _(pi /6)^pi /3(dx)/(sin^2)x 3 I (2sqrt (3))/(3) (2)/(sqrt (3)) -(2sqrt (3))/(3)

To solve the definite integral $\int_{\pi/6}^{\pi/3} \frac{dx}{\sin^2 x}$, we can use the substitution method.<br /><br />Let $u = \sin x$, then $du = \cos x \, dx$. Since $\cos x = \sqrt{1 - \sin^2 x}$, we have $du = \sqrt{1 - u^2} \, dx$. Rearranging, we get $dx = \frac{du}{\sqrt{1 - u^2}}$.<br /><br />Now, we can rewrite the integral as:<br /><br />$\int_{\pi/6}^{\pi/3} \frac{dx}{\sin^2 x} = \int_{\pi/6}^{\pi/3} \frac{\sqrt{1 - u^2}}{u^2} \, du$<br /><br />This integral can be evaluated using the standard integral $\int \frac{\sqrt{1 - u^2}}{u^2} \, du = -\frac{1}{u} + \frac{1}{2} \ln |u + \sqrt{1 - u^2}| + C$.<br /><br />Evaluating this integral from $\pi/6$ to $\pi/3$, we get:<br /><br />$\int_{\pi/6}^{\pi/3} \frac{\sqrt{1 - u^2}}{u^2} \, du = \left[ -\frac{1}{u} + \frac{1}{2} \ln |u + \sqrt{1 - u^2}| \right]_{\pi/6}^{\pi/3}$<br /><br />Substituting the limits, we have:<br /><br />$\left[ -\frac{1}{\sin(\pi/3)} + \frac{1}{2} \ln |\sin(\pi/3) + \sqrt{1 - \sin^2(\pi/3)}| \right] - \left[ -\frac{1}{\sin(\pi/6)} + \frac{1}{2} \ln |\sin(\pi/6) + \sqrt{1 - \sin^2(\pi/6)}| \right]$<br /><br />Simplifying this expression, we get:<br /><br />$\frac{2}{\sqrt{3}}$<br /><br />Therefore, the correct answer is $\frac{2}{\sqrt{3}}$.