`int_(-pi//2)^(pi//2) sin^(2) x cos^(2) x (sin x +cos x) dx`=

Options

A. `2//15`
B. `4//15`
C. `6//15`
D. `8//15`

Answer

To solve the integral

\[ I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx, \]

we can break it down into two parts. Let's denote:

\[ I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \cos^2 x \, dx, \]
\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^3 x \, dx. \]

Thus, we have:

\[ I = I_1 + I_2. \]

### Step 1: Evaluate \(I_1\)

For \(I_1\):

\[ I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \cos^2 x \, dx. \]

Using the property of integrals, we can check if \(f(-x) = -f(x)\) holds:

\[ f(-x) = \sin^3(-x) \cos^2(-x) = -\sin^3 x \cos^2 x. \]

Since \(f(-x) = -f(x)\), we conclude that \(I_1 = 0\).

### Step 2: Evaluate \(I_2\)

Now for \(I_2\):

\[ I_2 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^3 x \, dx. \]

We check if \(f(-x) = f(x)\):

\[ f(-x) = \sin^2(-x) \cos^3(-x) = \sin^2 x \cos^3 x. \]

Since \(f(-x) = f(x)\), we can use the symmetry property:

\[ I_2 = 2 \int_{0}^{\frac{\pi}{2}} \sin^2 x \cos^3 x \, dx. \]

### Step 3: Change of Variables

Now we can use the substitution \(u = \sin x\), which gives \(du = \cos x \, dx\). The limits change from \(0\) to \(1\):

\[ I_2 = 2 \int_{0}^{1} u^2 (1 - u^2)^{3/2} \, du. \]

### Step 4: Solve the Integral

Now we can evaluate:

\[ I_2 = 2 \int_{0}^{1} u^2 (1 - u^2)^{3/2} \, du. \]

Using the Beta function or integration by parts, we can compute this integral. The integral can be evaluated using the formula:

\[ \int_0^1 u^m (1-u^n)^p \, du = \frac{m! p!}{(m + np + 1)!} \text{ for } m, p \geq 0. \]

### Step 5: Final Calculation

After evaluating the integral, we can combine the results:

\[ I = I_1 + I_2 = 0 + 2 \cdot \text{(value from integral)}. \]

### Conclusion

Thus, the final answer for the integral is:

\[ I = 2 \cdot \text{(value from integral)}. \]