If `g(x) = int_(0)^(x) cost dt`, then `g(x+pi)` equals

Options

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

Answer

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

### Step 1: Write the expression for \( g(x + \pi) \)

We start by substituting \( x + \pi \) into the function \( g \):

\[ g(x + \pi) = \int_0^{x + \pi} \cos t \, dt \]

### Step 2: Split the integral

We can split the integral from \( 0 \) to \( x + \pi \) into two parts:

\[ g(x + \pi) = \int_0^{x} \cos t \, dt + \int_x^{x + \pi} \cos t \, dt \]

### Step 3: Evaluate the first integral

The first integral is simply \( g(x) \):

\[ \int_0^{x} \cos t \, dt = g(x) \]

### Step 4: Evaluate the second integral

Now, we need to evaluate the second integral \( \int_x^{x + \pi} \cos t \, dt \). We can use the property of the cosine function, which is periodic with a period of \( 2\pi \):

\[ \int_x^{x + \pi} \cos t \, dt = \int_0^{\pi} \cos t \, dt \]

### Step 5: Calculate \( \int_0^{\pi} \cos t \, dt \)

Now we compute this integral:

\[ \int_0^{\pi} \cos t \, dt = [\sin t]_0^{\pi} = \sin(\pi) - \sin(0) = 0 - 0 = 0 \]

### Step 6: Combine the results

Now we can combine the results from steps 3 and 5:

\[ g(x + \pi) = g(x) + \int_x^{x + \pi} \cos t \, dt = g(x) + 0 = g(x) \]

### Final Answer

Thus, we conclude that:

\[ g(x + \pi) = g(x) \]