The integral $$\int_{\pi/6}^{\pi/3} \sec^{2/3}x \cdot \text{cosec}^{4/3}x \, dx$$ is equal to

We have the definite integral

$$I=\int_{\pi/6}^{\pi/3}\sec^{2/3}x\;\cosec^{4/3}x\;dx.$$

First, we rewrite every trigonometric function in terms of $$\sin x$$ and $$\cos x$$. Remember that $$\sec x=\dfrac1{\cos x}$$ and $$\cosec x=\dfrac1{\sin x}$$. Substituting these definitions we get

$$I=\int_{\pi/6}^{\pi/3}\left(\dfrac1{\cos x}\right)^{2/3}\left(\dfrac1{\sin x}\right)^{4/3}dx.$$

Simplifying the powers gives

$$I=\int_{\pi/6}^{\pi/3}\cos^{-2/3}x\;\sin^{-4/3}x\;dx.$$

Now we introduce a substitution that will convert the integral into a simple power of the new variable. A very natural choice is the tangent function because its derivative contains $$\sec^2x$$, i.e. $$\dfrac1{\cos^2x}$$, and that will cancel the powers of $$\cos x$$ sitting in the integrand.

Let $$t=\tan x.$$ By differentiation,

$$\dfrac{dt}{dx}=\sec^2x=\dfrac1{\cos^2x}\quad\Longrightarrow\quad dx=\cos^2x\;dt.$$

We now change every factor in the integrand to the variable $$t$$. Starting with the powers:

$$$ \cos^{-2/3}x\;\sin^{-4/3}x\;dx =\cos^{-2/3}x\;\sin^{-4/3}x\;(\cos^2x\;dt) =\cos^{\,2-\frac23}x\;\sin^{-4/3}x\;dt =\cos^{\,\frac43}x\;\sin^{-4/3}x\;dt. $$$

Notice that $$\cos^{\,\frac43}x\;\sin^{-4/3}x=(\cos x/\sin x)^{4/3}=(\cot x)^{4/3}.$$ Since $$t=\tan x$$, we have $$\cot x=\dfrac1{\tan x}=\dfrac1t.$$ Therefore,

$$\cos^{\,\frac43}x\;\sin^{-4/3}x\;dt=(\cot x)^{4/3}dt=\left(\dfrac1t\right)^{4/3}dt=t^{-4/3}dt.$$

The integral is now

$$I=\int t^{-4/3}\,dt.$$

This is a pure power of $$t$$, so we integrate it with the usual power‐rule formula $$\displaystyle\int t^{n}\,dt=\dfrac{t^{n+1}}{n+1}+C$$, provided $$n\neq-1$$. Here $$n=-\dfrac43$$, so

$$$ \int t^{-4/3}\,dt=\frac{t^{-4/3+1}}{-4/3+1}+C =\frac{t^{-1/3}}{-1/3}+C =-3\,t^{-1/3}+C. $$$

Thus the antiderivative is $$-3\,\tan^{-1/3}x.$$

We now impose the original limits. When $$x=\pi/6$$, $$\tan(\pi/6)=\dfrac1{\sqrt3}$$. When $$x=\pi/3$$, $$\tan(\pi/3)=\sqrt3$$. Therefore,

$$$ I=\Bigl[-3\,\tan^{-1/3}x\Bigr]_{\,\pi/6}^{\,\pi/3} =-3\Bigl[\tan^{-1/3}(\pi/3)-\tan^{-1/3}(\pi/6)\Bigr]. $$$

We evaluate each term separately.

For $$x=\pi/3$$:

$$$ \tan^{-1/3}(\pi/3)=(\sqrt3)^{-1/3}=3^{-1/6}. $$$

For $$x=\pi/6$$:

$$$ \tan^{-1/3}(\pi/6)=\left(\frac1{\sqrt3}\right)^{-1/3}=(\sqrt3)^{1/3}=3^{1/6}. $$$

Substituting back gives

$$$ I=-3\Bigl[\,3^{-1/6}-3^{1/6}\Bigr] =3\Bigl[\,3^{1/6}-3^{-1/6}\Bigr]. $$$

We now distribute the factor $$3$$ through the bracket:

$$$ 3\cdot3^{1/6}=3^{1+1/6}=3^{7/6},\quad 3\cdot3^{-1/6}=3^{1-1/6}=3^{5/6}. $$$

Hence

$$I=3^{7/6}-3^{5/6}.$$

This expression matches Option A.

Hence, the correct answer is Option A.