If f(x)`=int_(0)^(x) sin^(4)t dt`, then `f(x+2pi)` is equal to

Options

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

Answer

To solve the problem, we need to find \( f(x + 2\pi) \) where \( f(x) = \int_0^x \sin^4 t \, dt \).

### Step-by-step Solution:

1. **Define the function**:
\[ f(x) = \int_0^x \sin^4 t \, dt \]

2. **Use the periodic property of sine**:
The sine function has a period of \( 2\pi \), which means:
\[ \sin^4(t + 2\pi) = \sin^4 t \]
Therefore, we can express \( f(x + 2\pi) \) as:
\[ f(x + 2\pi) = \int_0^{x + 2\pi} \sin^4 t \, dt \]

3. **Split the integral**:
We can split the integral into two parts:
\[ f(x + 2\pi) = \int_0^{x} \sin^4 t \, dt + \int_{x}^{x + 2\pi} \sin^4 t \, dt \]
The first part is simply \( f(x) \):
\[ f(x + 2\pi) = f(x) + \int_{x}^{x + 2\pi} \sin^4 t \, dt \]

4. **Evaluate the second integral**:
The integral \( \int_{x}^{x + 2\pi} \sin^4 t \, dt \) can be evaluated using the periodicity of the sine function:
\[ \int_{x}^{x + 2\pi} \sin^4 t \, dt = \int_0^{2\pi} \sin^4 t \, dt \]
This integral can be computed as follows:
\[ \int_0^{2\pi} \sin^4 t \, dt = 2 \int_0^{\pi} \sin^4 t \, dt \]
Using the reduction formula or known results, we find:
\[ \int_0^{\pi} \sin^4 t \, dt = \frac{3\pi}{8} \]
Thus,
\[ \int_0^{2\pi} \sin^4 t \, dt = 2 \cdot \frac{3\pi}{8} = \frac{3\pi}{4} \]

5. **Combine results**:
Now we can combine the results:
\[ f(x + 2\pi) = f(x) + \frac{3\pi}{4} \]

### Final Result:
\[ f(x + 2\pi) = f(x) + \frac{3\pi}{4} \]