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$\begin{aligned} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{\sin(2x)} & = \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{dx}{2\sin x \cos x} \quad \quad \small \color{#3D99F6}{\text{Let }y = \sin x \quad \Rightarrow dy = \cos x \space dx} \\ & = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{dy}{2\sin x \cos^2 x} \\ & = \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \frac{dy}{2 y (1 - y^2)} \\ & = \frac{1}{4} \int_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \left(\frac{1}{1-y} + \frac{2}{y} - \frac{1}{1+y} \right) dy \\ & = \frac{1}{4} \left[ -\ln(1-y) + 2 \ln y - \ln(1+y) \right]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\ & = \frac{1}{4} \left[ \ln \left( \frac {y^2}{1-y^2} \right) \right]_{\frac{1}{2}}^{\frac{\sqrt{3}}{2}} \\ & = \frac{1}{4} \left[ \ln \left(\frac{\frac{3}{4}}{1-\frac{3}{4}} \right) - \ln \left(\frac{\frac{1}{4}} {1-\frac{1}{4}} \right) \right] \\ & = \frac{1}{4} \left[ \ln 3 - \ln \left(\frac{1}{3} \right) \right] = \frac{\ln 9}{4} = \boxed{\ln \sqrt{3}} \end{aligned}$

$\sin(2x)$ can be represented as $\dfrac{2\tan x}{1+\tan^{2}x}$ .

$= \displaystyle \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{1+\tan^{2}x}{\tan x} \text{ d}x$ $= \displaystyle \frac{1}{2} \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \dfrac{\sec^{2}x}{\tan x} \text{ d}x$ Let $u=\tan x; \text{d}u=\sec^{2}x \text{ d}x$ $= \displaystyle \frac{1}{2} \int_{x=\frac{\pi}{6}}^{x=\frac{\pi}{3}} \dfrac{\text{d}u}{u}$ $= \displaystyle \frac{1}{2} \left[ \ln u \right]_{x=\frac{\pi}{6}}^{x=\frac{\pi}{3}} = \frac{1}{2} \left[ \ln \tan x \right]_{x=\frac{\pi}{6}}^{x=\frac{\pi}{3}}$ $= \displaystyle \dfrac{1}{2} \ln 3 = \boxed{\ln \sqrt{3}}$

$\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { 1 }{ \sin { 2x } } dx=\int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ \frac { { { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } } } }{ 2\sin { x } \cos { x } } dx } } \\ =-\frac { 1 }{ 2 } \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ -2\sin { x({ 2\cos { x) } }^{ -1 } } } +\frac { 1 }{ 2 } \int _{ \frac { \pi }{ 6 } }^{ \frac { \pi }{ 3 } }{ 2\cos { x({ 2\sin { x) } }^{ -1 } } } \\ =\begin{cases} \frac { \pi }{ 3 } \\ \frac { \pi }{ 6 } \end{cases}(\quad -\frac { 1 }{ 2 } \ln { \left| 2\cos { x } \right| } +\frac { 1 }{ 2 } \ln { 2\sin { x } } )\\ =\ln { \sqrt { 3 } }$

Clearly it is (sec(x))^2/tanx where substution tanx=t helps

For the mean value theorem, the solution is >0.52 and <0.609. Only the correct answer is in that range