The integral $\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{{{\sec }^{\dfrac{2}{3}}}x{{\csc }^{\dfrac{4}{3}}}xdx}$ is equal to
$\begin{align}
  & \left[ a \right]\ {{3}^{\dfrac{7}{6}}}-{{3}^{\dfrac{3}{6}}} \\
 & \left[ b \right]\,{{3}^{\dfrac{5}{3}}}-{{3}^{\dfrac{1}{3}}} \\
 & \left[ c \right]\,{{3}^{\dfrac{4}{3}}}-{{3}^{\dfrac{1}{3}}} \\
 & \left[ d \right]\,{{3}^{\dfrac{5}{6}}}-{{3}^{\dfrac{2}{3}}} \\
\end{align}$

Hint: Multiply and divide by ${{\sec }^{\dfrac{4}{3}}}x$. Hence prove that the integrand is equal to $\dfrac{{{\sec }^{2}}x}{{{\tan }^{\dfrac{4}{3}}}x}$. Put tanx = t and use the fact that the derivative of tanx is ${{\sec }^{2}}x$ and hence prove that the given integral is equal to $\int_{\dfrac{1}{\sqrt{3}}}^{\sqrt{3}}{\dfrac{dt}{{{t}^{\dfrac{4}{3}}}}}$. Use the fact that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1},n\ne -1$. Hence find the value of the given integral.

Complete step-by-step answer:
Let $I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{{{\sec }^{\dfrac{2}{3}}}x{{\csc }^{\dfrac{4}{3}}}xdx}$
Multiplying and dividing by ${{\sec }^{\dfrac{4}{3}}}x$, we get
$I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{{{\sec }^{\dfrac{2}{3}+\dfrac{4}{3}}}}x\dfrac{{{\csc }^{\dfrac{4}{3}}}x}{{{\sec }^{\dfrac{4}{3}}}x}dx$
We know that $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$.
Using the above result, we have
$I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{{{\sec }^{2}}}x{{\left( \dfrac{\csc x}{\sec x} \right)}^{\dfrac{4}{3}}}dx$
We know that $\dfrac{a}{b}=\dfrac{1}{\dfrac{b}{a}}$
Using the above result, we have
$I=\int_{\dfrac{\pi }{6}}^{\dfrac{\pi }{3}}{{{\sec }^{2}}}x{{\left( \dfrac{1}{\dfrac{\sec x}{\csc x}} \right)}^{\dfrac{4}{3}}}dx$
We know that $\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\sec x}{\csc x}$
Using the above result, we get
$I=\int_{\pi /6}^{\pi /3}{\dfrac{{{\sec }^{2}}x}{{{\tan }^{\dfrac{4}{3}}}x}dx}$
Put tanx = t
Differentiating both sides, we get
${{\sec }^{2}}xdx=dt$
When $x=\dfrac{\pi }{3},t=\sqrt{3}$
When $x=\dfrac{\pi }{6},t=\dfrac{1}{\sqrt{3}}$
Hence, we have
$I=\int_{1/\sqrt{3}}^{\sqrt{3}}{\dfrac{dt}{{{t}^{\dfrac{4}{3}}}}=}\int_{1/\sqrt{3}}^{\sqrt{3}}{{{t}^{\dfrac{-4}{3}}}dt}$
We know that $\int{{{x}^{n}}dx}=\dfrac{{{x}^{n+1}}}{n+1},n\ne -1$.
We know according to fundamental theorem of calculus that if $\int{f\left( x \right)dx}=F\left( x \right)$ then $\int\limits_{a}^{b}{f\left( x \right)dx}=F\left( b \right)-F\left( a \right)$
Hence, we have
\[I=\left. \left( -\dfrac{3}{\sqrt[3]{t}} \right) \right|_{1/\sqrt{3}}^{\sqrt{3}}={{3}^{\dfrac{7}{6}}}-{{3}^{\dfrac{5}{6}}}\]

So, the correct answer is “Option A”.

Note: [1] A general trick for solving questions of the form $\int{{{\sec }^{a}}x{{\csc }^{2-a}}x}$ is by dividing and multiplying both sides by ${{\sec }^{2-a}}x$ and put tanx = t. The integral reduces to $\int{\dfrac{dt}{{{t}^{2-a}}}}$ which can be easily solved. This question is also of this form and hence this method is applied.
[2] It is generally useful to change the limits when changing the variable of integration as this avoids the process of reverting to the original variable and hence minimizes chances of making calculation mistakes.