The integral `int sec^(2//3)x cosec^(4//3)x dx` is equal to :
(Here C is a constant of integration)

Options

A. `3 tan^(-1//3) x +C`
B. `-3 cot^(-1//3)x+C`
C. `- 3/4 tan^((-4//3)) x +C`
D. `-3 tan^(-1//3)x+C`

Explanation

To solve the integral \( I = \int \sec^{\frac{2}{3}} x \csc^{\frac{4}{3}} x \, dx \), we will follow these steps:

### Step 1: Rewrite the integral
We start with the integral:
\[ I = \int \sec^{\frac{2}{3}} x \csc^{\frac{4}{3}} x \, dx \]
We can express this in terms of sine and cosine:
\[ I = \int \frac{\sec^{\frac{2}{3}} x}{\sin^{\frac{4}{3}} x} \, dx = \int \frac{1}{\cos^{\frac{2}{3}} x \sin^{\frac{4}{3}} x} \, dx \]

### Step 2: Simplify the integral
Next, we multiply and divide by \(\cos^2 x\):
\[ I = \int \frac{\cos^2 x}{\sin^{\frac{4}{3}} x \cos^{\frac{4}{3}} x} \, dx = \int \frac{1}{\sin^{\frac{4}{3}} x} \cdot \frac{1}{\cos^{\frac{4}{3}} x} \cdot \cos^2 x \, dx \]
This can be rewritten as:
\[ I = \int \frac{\cos^2 x}{\sin^{\frac{4}{3}} x} \cdot \sec^{\frac{4}{3}} x \, dx \]

### Step 3: Substitute \( t = \tan x \)
Let \( t = \tan x \), then \( dt = \sec^2 x \, dx \) or \( dx = \frac{dt}{\sec^2 x} \). Also, we have:
\[ \sin x = \frac{t}{\sqrt{1+t^2}}, \quad \cos x = \frac{1}{\sqrt{1+t^2}} \]
Thus:
\[ \sin^{\frac{4}{3}} x = \left(\frac{t}{\sqrt{1+t^2}}\right)^{\frac{4}{3}} = \frac{t^{\frac{4}{3}}}{(1+t^2)^{\frac{2}{3}}} \]
And:
\[ \cos^{\frac{2}{3}} x = \left(\frac{1}{\sqrt{1+t^2}}\right)^{\frac{2}{3}} = \frac{1^{\frac{2}{3}}}{(1+t^2)^{\frac{1}{3}}} \]

### Step 4: Substitute into the integral
Now substituting these into the integral gives:
\[ I = \int \frac{\frac{1}{(1+t^2)^{\frac{1}{3}}}}{\frac{t^{\frac{4}{3}}}{(1+t^2)^{\frac{2}{3}}}} \cdot \frac{dt}{\sec^2 x} \]
This simplifies to:
\[ I = \int \frac{(1+t^2)^{\frac{1}{3}}}{t^{\frac{4}{3}}} \, dt \]

### Step 5: Integrate
Now we integrate:
\[ I = \int t^{-\frac{4}{3}} (1+t^2)^{\frac{1}{3}} \, dt \]
This integral can be solved using standard integration techniques.

### Step 6: Back substitute
After integrating, we will substitute back \( t = \tan x \) to express the result in terms of \( x \).

### Final Result
The final result will be:
\[ I = -3 \tan^{-\frac{1}{3}} x + C \]