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

Options

A. `(15)/(4)`
B. `(4)/(15)`
C. `(5)/(16)`
D. `(16)/(5)`

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 will use the property of definite integrals that states:

\[ \int_a^b f(x) \, dx = \int_a^b f(a + b - x) \, dx. \]

### Step 1: Apply the property of definite integrals

Let’s denote the integral as:

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

Using the property, we substitute \( x \) with \( -x \):

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

### Step 2: Simplify the expression

Using the properties of sine and cosine:

- \(\sin(-x) = -\sin x\)
- \(\cos(-x) = \cos x\)

We can rewrite the integral:

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

### Step 3: Set up the second equation

Now we have two expressions for \( I \):

1. \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (\sin x + \cos x) \, dx \) (Equation 1)
2. \( I = \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sin^2 x \cos^2 x (-\sin x + \cos x) \, dx \) (Equation 2)

### Step 4: Add the two equations

Adding both equations:

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

This simplifies to:

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

### Step 5: Factor out the constant

We can factor out the 2:

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

### Step 6: Change of variables

Let \( t = \sin x \), then \( dt = \cos x \, dx \).

When \( x = -\frac{\pi}{2} \), \( t = -1 \) and when \( x = \frac{\pi}{2} \), \( t = 1 \).

Thus, the integral becomes:

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

### Step 7: Expand and integrate

Expanding the integrand:

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

Now, we can integrate term by term:

\[ I = \left[ \frac{t^3}{3} - \frac{t^5}{5} \right]_{-1}^{1}. \]

### Step 8: Evaluate the limits

Calculating at the limits:

\[ I = \left( \frac{1^3}{3} - \frac{1^5}{5} \right) - \left( \frac{(-1)^3}{3} - \frac{(-1)^5}{5} \right). \]

This simplifies to:

\[ I = \left( \frac{1}{3} - \frac{1}{5} \right) - \left( -\frac{1}{3} + \frac{1}{5} \right). \]

### Step 9: Combine the results

Combining these gives:

\[ I = \frac{1}{3} - \frac{1}{5} + \frac{1}{3} - \frac{1}{5} = \frac{2}{3} - \frac{2}{5}. \]

Finding a common denominator (15):

\[ I = \frac{10}{15} - \frac{6}{15} = \frac{4}{15}. \]

### Final Result

Thus, the value of the integral is:

\[ I = \frac{4}{15}. \]