The value of `int_(-pi//2)^(pi//2)( 3sinx+sin^(3)x)dx` is

Options

A. 3
B. 2
C. 0
D. `10/3`

Answer

To solve the integral \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (3 \sin x + \sin^3 x) \, dx \), we can use the property of definite integrals that involves the symmetry of the integrand.

### Step 1: Identify the integrand
The integrand is \( f(x) = 3 \sin x + \sin^3 x \).

### Step 2: Use the property of definite integrals
We can use the property that states:
\[ I = \int_{-a}^{a} f(x) \, dx = \int_{-a}^{a} \left( f(x) + f(-x) \right) \, dx \]
for odd functions, where \( f(-x) = -f(x) \).

### Step 3: Calculate \( f(-x) \)
Now, we calculate \( f(-x) \):
\[ f(-x) = 3 \sin(-x) + \sin^3(-x) = 3(-\sin x) + (-\sin x)^3 = -3 \sin x - \sin^3 x \]

### Step 4: Add \( f(x) \) and \( f(-x) \)
Now, we can add \( f(x) \) and \( f(-x) \):
\[ f(x) + f(-x) = (3 \sin x + \sin^3 x) + (-3 \sin x - \sin^3 x) = 0 \]

### Step 5: Conclude the integral
Since \( f(x) + f(-x) = 0 \), we have:
\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (3 \sin x + \sin^3 x) \, dx = 0 \]

### Final Answer
Thus, the value of the integral is:
\[ \boxed{0} \]