If `h(x)=int_(0)^(x)sin^(4)tdt`, then `h(x+pi)` equals

Options

A. `(h(x))/(h(pi))`
B. `h(x).h(pi)`
C. `h(x)-h(pi)`
D. `h(x)+h(pi)`

Answer

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

### Step-by-Step Solution:

1. **Define \( h(x + \pi) \)**:
\[ h(x + \pi) = \int_0^{x + \pi} \sin^4 t \, dt \]

2. **Break the Integral**:
We can break the integral from \( 0 \) to \( x + \pi \) into two parts:
\[ h(x + \pi) = \int_0^{\pi} \sin^4 t \, dt + \int_{\pi}^{x + \pi} \sin^4 t \, dt \]

3. **Evaluate the First Integral**:
The first part is simply:
\[ \int_0^{\pi} \sin^4 t \, dt \]
This integral can be computed separately, but we will denote it as \( I \) for now.

4. **Change of Variables for the Second Integral**:
For the second integral, we can use the substitution \( u = t - \pi \). Then \( dt = du \) and when \( t = \pi \), \( u = 0 \) and when \( t = x + \pi \), \( u = x \):
\[ \int_{\pi}^{x + \pi} \sin^4 t \, dt = \int_0^{x} \sin^4(u + \pi) \, du \]
Since \( \sin(u + \pi) = -\sin u \), we have:
\[ \sin^4(u + \pi) = (-\sin u)^4 = \sin^4 u \]
Thus, the integral becomes:
\[ \int_{\pi}^{x + \pi} \sin^4 t \, dt = \int_0^{x} \sin^4 u \, du = h(x) \]

5. **Combine the Results**:
Now we can combine the results:
\[ h(x + \pi) = I + h(x) \]
where \( I = \int_0^{\pi} \sin^4 t \, dt \).

6. **Final Expression**:
Therefore, we have:
\[ h(x + \pi) = \int_0^{\pi} \sin^4 t \, dt + h(x) \]

### Conclusion:
The final result is:
\[ h(x + \pi) = h(x) + \int_0^{\pi} \sin^4 t \, dt \]