The value of `int_(a)^(a+(pi//2))(sin^(4)x+cos^(4)x)dx` is

Options

A. `(3pi)/(8)`
B. `a((pi)/(2))^(2)`
C. `(3pia^(2))/(8)`
D. none of these

Answer

To solve the integral \( I = \int_{a}^{a + \frac{\pi}{2}} \left( \sin^4 x + \cos^4 x \right) dx \), we will follow these steps:

### Step 1: Simplify the integrand
We can use the identity \( \sin^4 x + \cos^4 x = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x \). Since \( \sin^2 x + \cos^2 x = 1 \), we have:
\[ \sin^4 x + \cos^4 x = 1 - 2\sin^2 x \cos^2 x \]

### Step 2: Rewrite the integral
Substituting this back into the integral, we get:
\[ I = \int_{a}^{a + \frac{\pi}{2}} \left( 1 - 2\sin^2 x \cos^2 x \right) dx \]

### Step 3: Use the double angle identity
Recall that \( \sin^2 x \cos^2 x = \frac{1}{4} \sin^2(2x) \). Therefore, we can rewrite the integral as:
\[ I = \int_{a}^{a + \frac{\pi}{2}} \left( 1 - \frac{1}{2} \sin^2(2x) \right) dx \]

### Step 4: Split the integral
Now we can split the integral into two parts:
\[ I = \int_{a}^{a + \frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{a}^{a + \frac{\pi}{2}} \sin^2(2x) \, dx \]

### Step 5: Evaluate the first integral
The first integral is straightforward:
\[ \int_{a}^{a + \frac{\pi}{2}} 1 \, dx = \left[ x \right]_{a}^{a + \frac{\pi}{2}} = \left( a + \frac{\pi}{2} \right) - a = \frac{\pi}{2} \]

### Step 6: Evaluate the second integral
For the second integral, we use the identity \( \sin^2(2x) = \frac{1 - \cos(4x)}{2} \):
\[ \int_{a}^{a + \frac{\pi}{2}} \sin^2(2x) \, dx = \int_{a}^{a + \frac{\pi}{2}} \frac{1 - \cos(4x)}{2} \, dx = \frac{1}{2} \int_{a}^{a + \frac{\pi}{2}} 1 \, dx - \frac{1}{2} \int_{a}^{a + \frac{\pi}{2}} \cos(4x) \, dx \]
The first part evaluates to:
\[ \frac{1}{2} \cdot \frac{\pi}{2} = \frac{\pi}{4} \]
The second part evaluates as:
\[ \int \cos(4x) \, dx = \frac{1}{4} \sin(4x) \quad \text{so} \quad \left[ \frac{1}{4} \sin(4x) \right]_{a}^{a + \frac{\pi}{2}} = \frac{1}{4} \left( \sin(2\pi) - \sin(4a) \right) = -\frac{1}{4} \sin(4a) \]
Thus, combining these results:
\[ \int_{a}^{a + \frac{\pi}{2}} \sin^2(2x) \, dx = \frac{\pi}{4} + \frac{1}{4} \sin(4a) \]

### Step 7: Substitute back into the integral
Now substituting back into \( I \):
\[ I = \frac{\pi}{2} - \frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{4} \sin(4a) \right) = \frac{\pi}{2} - \frac{\pi}{8} - \frac{1}{8} \sin(4a) \]
This simplifies to:
\[ I = \frac{4\pi}{8} - \frac{\pi}{8} - \frac{1}{8} \sin(4a) = \frac{3\pi}{8} - \frac{1}{8} \sin(4a) \]

### Final Result
Thus, the value of the integral is:
\[ I = \frac{3\pi}{8} - \frac{1}{8} \sin(4a) \]