Evaluate: `int_(-pi/2)^(pi/2)sin^2xcos^2x(sinx+cosx)dx`

Explanation

To evaluate the integral

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

we can start by expanding the integrand:

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

Let's denote these two integrals as:

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

Thus, we have:

\[ I = I_1 + I_2. \]

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

The function \(\sin^3 x \cos^2 x\) is an odd function because \(\sin^3(-x) = -\sin^3(x)\) and \(\cos^2(-x) = \cos^2(x)\). Therefore, the integral of an odd function over symmetric limits is zero:

\[ I_1 = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^3 x \cos^2 x \, dx = 0. \]

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

Now we consider \(I_2\):

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

The function \(\sin^2 x \cos^3 x\) is an even function because \(\sin^2(-x) = \sin^2(x)\) and \(\cos^3(-x) = -\cos^3(x)\). Thus, we can simplify the integral:

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

### Step 3: Use substitution

Now we can use the substitution \(t = \sin x\), which gives \(dt = \cos x \, dx\). The limits change from \(x = 0\) to \(x = \frac{\pi}{2}\), which corresponds to \(t = 0\) to \(t = 1\). Thus:

\[ I_2 = 2 \int_{0}^{1} t^2 (1 - t^2) \, dt, \]

where we used \(\sin^2 x = t^2\) and \(\cos^3 x = (1 - t^2)^{3/2}\).

### Step 4: Evaluate the integral

Now we can expand the integrand:

\[ I_2 = 2 \int_{0}^{1} (t^2 - t^4) \, dt. \]

Calculating the integral:

\[ = 2 \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_{0}^{1} = 2 \left( \frac{1}{3} - \frac{1}{5} \right). \]

Finding a common denominator (15):

\[ = 2 \left( \frac{5}{15} - \frac{3}{15} \right) = 2 \cdot \frac{2}{15} = \frac{4}{15}. \]

### Final Result

Thus, we have:

\[ I = I_1 + I_2 = 0 + \frac{4}{15} = \frac{4}{15}. \]

So, the final answer is:

\[ \boxed{\frac{4}{15}}. \]