If `f (x) = int_(0) ^(x) (cos ^(4) t + sin^(4) t ) dt, f (x + pi) ` will be equal to

Options

A. `f (x) + f (pi)`
B. ` f (x) + 2 f(pi)`
C. ` f (x) + f ((pi)/(2))`
D. ` f (x) + 2f ((pi)/(2))`

Answer

To solve the problem, we need to find \( f(x + \pi) \) given that \( f(x) = \int_0^x (\cos^4 t + \sin^4 t) \, dt \).

### Step-by-Step Solution:

1. **Understanding the Function**:
We have:
\[ f(x) = \int_0^x (\cos^4 t + \sin^4 t) \, dt \]
We need to evaluate \( f(x + \pi) \).

2. **Substituting into the Function**:
We can express \( f(x + \pi) \) as:
\[ f(x + \pi) = \int_0^{x + \pi} (\cos^4 t + \sin^4 t) \, dt \]

3. **Breaking the Integral**:
We can break the integral into two parts:
\[ f(x + \pi) = \int_0^{\pi} (\cos^4 t + \sin^4 t) \, dt + \int_{\pi}^{x + \pi} (\cos^4 t + \sin^4 t) \, dt \]
The first integral is \( f(\pi) \):
\[ f(x + \pi) = f(\pi) + \int_{\pi}^{x + \pi} (\cos^4 t + \sin^4 t) \, dt \]

4. **Changing Variables in the Second Integral**:
For the second integral, we can make a substitution. Let \( u = t - \pi \), then \( dt = du \) and when \( t = \pi \), \( u = 0 \) and when \( t = x + \pi \), \( u = x \):
\[ \int_{\pi}^{x + \pi} (\cos^4 t + \sin^4 t) \, dt = \int_0^{x} (\cos^4 (u + \pi) + \sin^4 (u + \pi)) \, du \]

5. **Using Trigonometric Identities**:
Using the identities:
\[ \cos(u + \pi) = -\cos u \quad \text{and} \quad \sin(u + \pi) = -\sin u \]
We have:
\[ \cos^4(u + \pi) = (-\cos u)^4 = \cos^4 u \quad \text{and} \quad \sin^4(u + \pi) = (-\sin u)^4 = \sin^4 u \]
Thus:
\[ \int_{\pi}^{x + \pi} (\cos^4 t + \sin^4 t) \, dt = \int_0^{x} (\cos^4 u + \sin^4 u) \, du = f(x) \]

6. **Combining Results**:
Now we can combine our results:
\[ f(x + \pi) = f(\pi) + f(x) \]

### Conclusion:
Thus, the final result is:
\[ f(x + \pi) = f(x) + f(\pi) \]