Let `f (x) = int x ^(2) cos ^(2)x (2x + 6 tan x - 2x tan ^(2) x ) dx and f (x)` passes through the point `(pi, 0)`
The number of solution (s) of the equation `f (x) =x ^(3)` in `[0, 2pi]` be:

Options

A. 0
B. 3
C. 2
D. None of these

Answer

To solve the problem, we need to evaluate the integral and determine the number of solutions to the equation \( f(x) = x^3 \) in the interval \([0, 2\pi]\). Let's break down the steps:

### Step 1: Define the function
We start with the function given in the problem:
\[ f(x) = \int x^2 \cos^2 x (2x + 6 \tan x - 2x \tan^2 x) \, dx \]

### Step 2: Simplify the integrand
We can simplify the integrand by distributing \( x^2 \cos^2 x \):
\[ f(x) = \int \left( 2x^3 \cos^2 x + 6x^2 \sin x \cos x - 2x^3 \sin^2 x \right) \, dx \]

### Step 3: Rewrite the integral
We can rewrite the integral as:
\[ f(x) = \int \left( 2x^3 \cos^2 x - 2x^3 \sin^2 x + 6x^2 \sin x \cos x \right) \, dx \]
This can be simplified further:
\[ f(x) = \int \left( 2x^3 (\cos^2 x - \sin^2 x) + 6x^2 \sin x \cos x \right) \, dx \]

### Step 4: Integrate each term
Now we will integrate each term separately:
1. For the first term \( 2x^3 (\cos^2 x - \sin^2 x) \), we can use integration by parts.
2. For the second term \( 6x^2 \sin x \cos x \), we can use the identity \( \sin(2x) = 2 \sin x \cos x \).

### Step 5: Apply integration by parts
Let’s denote:
- \( u = x^3 \) and \( dv = 2(\cos^2 x - \sin^2 x) \, dx \)
Using integration by parts:
\[ f(x) = x^3 \int 2(\cos^2 x - \sin^2 x) \, dx - \int \left( \int 2(\cos^2 x - \sin^2 x) \, dx \right) \cdot 3x^2 \, dx \]

### Step 6: Combine results
After integrating both parts, we combine the results to get:
\[ f(x) = x^3 \sin(2x) + C \]

### Step 7: Determine the constant \( C \)
Since \( f(\pi) = 0 \), we can substitute \( x = \pi \):
\[ f(\pi) = \pi^3 \sin(2\pi) + C = 0 \implies C = 0 \]
Thus, \( f(x) = x^3 \sin(2x) \).

### Step 8: Set up the equation
Now we need to solve the equation:
\[ f(x) = x^3 \implies x^3 \sin(2x) = x^3 \]
This simplifies to:
\[ x^3 (\sin(2x) - 1) = 0 \]
The solutions occur when:
1. \( x^3 = 0 \) which gives \( x = 0 \).
2. \( \sin(2x) - 1 = 0 \) which gives \( 2x = \frac{\pi}{2} + 2k\pi \) for \( k \in \mathbb{Z} \).

### Step 9: Find solutions in \([0, 2\pi]\)
From \( 2x = \frac{\pi}{2} + 2k\pi \):
- For \( k = 0 \): \( x = \frac{\pi}{4} \)
- For \( k = 1 \): \( x = \frac{5\pi}{4} \)
- For \( k = 2 \): \( x = \frac{9\pi}{4} \) (not in \([0, 2\pi]\))

Thus, the solutions in \([0, 2\pi]\) are \( x = 0, \frac{\pi}{4}, \frac{5\pi}{4} \).

### Conclusion
The total number of solutions to the equation \( f(x) = x^3 \) in the interval \([0, 2\pi]\) is **3**.

---