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 evaluate each term separately and analyze their contributions to the integral.

### Step 1: Evaluate \( I_1 = \int_{-a}^{a} p \tan^3 x \, dx \)

1. **Recognize the odd function**: The function \( \tan^3 x \) is an odd function because \( \tan(-x) = -\tan(x) \). Therefore, \( \tan^3(-x) = -\tan^3(x) \).
2. **Integral of an odd function**: The integral of an odd function over a symmetric interval around zero is zero.
\[ I_1 = p \int_{-a}^{a} \tan^3 x \, dx = 0 \]

### Step 2: Evaluate \( I_2 = \int_{-a}^{a} q \cos^2 x \, dx \)

1. **Recognize the even function**: The function \( \cos^2 x \) is an even function because \( \cos^2(-x) = \cos^2(x) \).
2. **Use the identity**: We can use the identity \( \cos^2 x = \frac{1 + \cos(2x)}{2} \).
\[ I_2 = q \int_{-a}^{a} \cos^2 x \, dx = q \int_{-a}^{a} \frac{1 + \cos(2x)}{2} \, dx \]
3. **Separate the integral**:
\[ I_2 = q \left( \frac{1}{2} \int_{-a}^{a} 1 \, dx + \frac{1}{2} \int_{-a}^{a} \cos(2x) \, dx \right) \]
4. **Evaluate the integrals**:
- The first integral: \( \int_{-a}^{a} 1 \, dx = 2a \).
- The second integral: The integral of \( \cos(2x) \) over a symmetric interval is zero.
\[ I_2 = q \left( \frac{1}{2} (2a) + 0 \right) = qa \]

### Step 3: Evaluate \( I_3 = \int_{-a}^{a} r \sin x \, dx \)

1. **Recognize the odd function**: The function \( \sin x \) is also an odd function because \( \sin(-x) = -\sin(x) \).
2. **Integral of an odd function**: The integral of an odd function over a symmetric interval around zero is zero.
\[ I_3 = r \int_{-a}^{a} \sin x \, dx = 0 \]

### Final Calculation of \( I \)

Combining the results:
\[ I = I_1 + I_2 + I_3 = 0 + qa + 0 = qa \]

### Conclusion

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