The maximum value of f(x)`=int_(0)^(1) t sin (x+pi t)dt` is

Options

A. `(1)/(pi)sqrt(pi^(2)+4)`
B. `(1)/(pi^(2))sqrt(pi^(2)+4)`
C. `sqrt(pi^(2)+4)`
D. `(1)/(2pi^(2))sqrt(pi^(2)+4)`

Explanation

To find the maximum value of the function \( f(x) = \int_0^1 t \sin(x + \pi t) \, dt \), we will follow these steps:

### Step 1: Evaluate the integral
We start by evaluating the integral \( f(x) \):
\[ f(x) = \int_0^1 t \sin(x + \pi t) \, dt \]

### Step 2: Use integration by parts
We will use integration by parts, where we let:
- \( u = t \) and \( dv = \sin(x + \pi t) \, dt \)

Then, we differentiate and integrate:
- \( du = dt \)
- \( v = -\frac{1}{\pi} \cos(x + \pi t) \)

Using integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ f(x) = \left[ -\frac{t}{\pi} \cos(x + \pi t) \right]_0^1 + \frac{1}{\pi} \int_0^1 \cos(x + \pi t) \, dt \]

### Step 3: Evaluate the boundary terms
Evaluating the boundary terms:
\[ \left[ -\frac{t}{\pi} \cos(x + \pi t) \right]_0^1 = -\frac{1}{\pi} \cos(x + \pi) + 0 = -\frac{1}{\pi} (-\cos x) = \frac{1}{\pi} \cos x \]

### Step 4: Evaluate the remaining integral
Now we need to evaluate the integral \( \int_0^1 \cos(x + \pi t) \, dt \):
\[ \int_0^1 \cos(x + \pi t) \, dt = \left[ \frac{1}{\pi} \sin(x + \pi t) \right]_0^1 = \frac{1}{\pi} (\sin(x + \pi) - \sin(x)) = \frac{1}{\pi} (-\sin x - \sin x) = -\frac{2}{\pi} \sin x \]

### Step 5: Combine results
Putting it all together:
\[ f(x) = \frac{1}{\pi} \cos x - \frac{2}{\pi^2} \sin x \]

### Step 6: Find maximum value
To find the maximum value of \( f(x) \), we express it in the form \( A \cos x + B \sin x \):
\[ f(x) = \frac{1}{\pi} \cos x - \frac{2}{\pi^2} \sin x \]
where \( A = \frac{1}{\pi} \) and \( B = -\frac{2}{\pi^2} \).

The maximum value of \( A \cos x + B \sin x \) is given by:
\[ \sqrt{A^2 + B^2} \]
Calculating:
\[ A^2 = \left(\frac{1}{\pi}\right)^2 = \frac{1}{\pi^2} \]
\[ B^2 = \left(-\frac{2}{\pi^2}\right)^2 = \frac{4}{\pi^4} \]
Thus,
\[ A^2 + B^2 = \frac{1}{\pi^2} + \frac{4}{\pi^4} = \frac{\pi^4 + 4}{\pi^4} \]
Taking the square root:
\[ \sqrt{A^2 + B^2} = \frac{\sqrt{\pi^4 + 4}}{\pi^2} \]

### Final Answer
The maximum value of \( f(x) \) is:
\[ \frac{\sqrt{\pi^4 + 4}}{\pi^2} \]