Let `y = f(x)` be a differentiable curve satisfying `int_(2)^(x)f(t)dt=(x^(2))/(2)+int_(x)^(2)t^(2)f(t)dt`, then `int_(-pi//4)^(pi//4)(f(x)+x^(9)-x^(3)+x+1)/(cos^(2)x)dx` equals :

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem step by step, we will first analyze the given equation and find the function \( f(x) \).

### Step 1: Analyze the given equation
We have the equation:
\[ \int_{2}^{x} f(t) dt = \frac{x^2}{2} + \int_{x}^{2} t^2 f(t) dt \]
This equation relates the integral of \( f(t) \) from 2 to \( x \) with a quadratic term and another integral.

### Step 2: Differentiate both sides with respect to \( x \)
Using the Fundamental Theorem of Calculus and Leibniz's rule, we differentiate both sides:
\[ \frac{d}{dx} \left( \int_{2}^{x} f(t) dt \right) = f(x) \]
For the right side:
\[ \frac{d}{dx} \left( \frac{x^2}{2} \right) + \frac{d}{dx} \left( \int_{x}^{2} t^2 f(t) dt \right) \]
Using Leibniz's rule on the second term:
\[ \frac{d}{dx} \left( \int_{x}^{2} t^2 f(t) dt \right) = -x^2 f(x) \]
Thus, we have:
\[ f(x) = x + 0 - x^2 f(x) \]
This simplifies to:
\[ f(x) + x^2 f(x) = x \]

### Step 3: Solve for \( f(x) \)
Factoring out \( f(x) \):
\[ f(x)(1 + x^2) = x \]
Therefore,
\[ f(x) = \frac{x}{1 + x^2} \]

### Step 4: Set up the integral to evaluate
Now we need to evaluate:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{f(x) + x^9 - x^3 + x + 1}{\cos^2 x} dx \]
Substituting \( f(x) \):
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\frac{x}{1 + x^2} + x^9 - x^3 + x + 1}{\cos^2 x} dx \]

### Step 5: Identify odd and even functions
We analyze the integrand:
1. \( \frac{x}{1 + x^2} \) is an odd function.
2. \( x^9 \) is an odd function.
3. \( -x^3 \) is an odd function.
4. \( x \) is an odd function.
5. \( 1 \) is an even function.

Combining these, the sum of the odd functions results in an odd function, while the even function remains unchanged.

### Step 6: Evaluate the integral
Since the integral of an odd function over a symmetric interval around zero is zero:
\[ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \left( \frac{x}{1 + x^2} + x^9 - x^3 + x \right) \frac{1}{\cos^2 x} dx = 0 \]
Thus, we only need to evaluate:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{1}{\cos^2 x} dx \]
This is:
\[ I = \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec^2 x \, dx \]

### Step 7: Calculate the integral of \( \sec^2 x \)
The integral of \( \sec^2 x \) is:
\[ \int \sec^2 x \, dx = \tan x \]
Evaluating from \( -\frac{\pi}{4} \) to \( \frac{\pi}{4} \):
\[ I = \tan\left(\frac{\pi}{4}\right) - \tan\left(-\frac{\pi}{4}\right) = 1 - (-1) = 2 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{2} \]