Let `I=int_(-a)^(a) (p tan^(3) x + q cos^(2)x + r sin x)dx`, where p, q, r are arbitrary constants. The numerical value of I depends on:

Options

A. p, q, r, a
B. q, r, a
C. q, a
D. p, r, a

Answer

To solve the integral \( I = \int_{-a}^{a} (p \tan^3 x + q \cos^2 x + r \sin x) \, dx \), we will analyze the integrand to determine which terms contribute to the value of the integral.

### Step 1: Analyze the integrand
The integrand consists of three terms: \( p \tan^3 x \), \( q \cos^2 x \), and \( r \sin x \). We need to determine whether each term is an even function or an odd function.

### Step 2: Determine the nature of each term
1. **For \( p \tan^3 x \)**:
- Check if it is odd or even:
\[ f_1(-x) = \tan^3(-x) = (-\tan x)^3 = -\tan^3 x = -f_1(x) \]
- Conclusion: \( p \tan^3 x \) is an odd function.

2. **For \( q \cos^2 x \)**:
- Check if it is odd or even:
\[ f_2(-x) = \cos^2(-x) = \cos^2 x = f_2(x) \]
- Conclusion: \( q \cos^2 x \) is an even function.

3. **For \( r \sin x \)**:
- Check if it is odd or even:
\[ f_3(-x) = \sin(-x) = -\sin x = -f_3(x) \]
- Conclusion: \( r \sin x \) is an odd function.

### Step 3: Apply properties of definite integrals
Using the properties of definite integrals:
- The integral of an odd function over a symmetric interval around zero is zero.
- The integral of an even function can be simplified as:
\[ \int_{-a}^{a} f(x) \, dx = 2 \int_{0}^{a} f(x) \, dx \]

### Step 4: Evaluate the integral
Now we can break down the integral \( I \):
\[ I = \int_{-a}^{a} (p \tan^3 x) \, dx + \int_{-a}^{a} (q \cos^2 x) \, dx + \int_{-a}^{a} (r \sin x) \, dx \]
- The first term \( \int_{-a}^{a} (p \tan^3 x) \, dx = 0 \) (odd function).
- The second term \( \int_{-a}^{a} (q \cos^2 x) \, dx = 2 \int_{0}^{a} (q \cos^2 x) \, dx \) (even function).
- The third term \( \int_{-a}^{a} (r \sin x) \, dx = 0 \) (odd function).

Thus, we have:
\[ I = 0 + 2 \int_{0}^{a} (q \cos^2 x) \, dx + 0 = 2 \int_{0}^{a} (q \cos^2 x) \, dx \]

### Step 5: Simplify the integral
Now, we can compute the integral:
\[ I = 2q \int_{0}^{a} \cos^2 x \, dx \]
Using the identity \( \cos^2 x = \frac{1 + \cos 2x}{2} \):
\[ I = 2q \int_{0}^{a} \frac{1 + \cos 2x}{2} \, dx = q \int_{0}^{a} (1 + \cos 2x) \, dx \]
Calculating the integral:
\[ = q \left[ x + \frac{\sin 2x}{2} \right]_{0}^{a} = q \left( a + \frac{\sin 2a}{2} - 0 \right) = q \left( a + \frac{\sin 2a}{2} \right) \]

### Conclusion
The numerical value of \( I \) depends on \( q \) and \( a \). Therefore, the answer is:

**The numerical value of \( I \) depends on \( q \) and \( a \).**