`" If " f(x) =int_(0)^(x)" t sin t dt tehn " f(x) is `

Explanation

To solve the problem, we need to find the function \( f(x) \) defined as:

\[ f(x) = \int_{0}^{x} t \sin t \, dt \]

We will use integration by parts to evaluate this integral.

### Step 1: Identify \( u \) and \( dv \)
We choose:
- \( u = t \) (which means \( du = dt \))
- \( dv = \sin t \, dt \) (which means \( v = -\cos t \))

### Step 2: Apply Integration by Parts
Using the integration by parts formula:
\[ \int u \, dv = uv - \int v \, du \]
we substitute our choices:
\[ f(x) = \left[ t (-\cos t) \right]_{0}^{x} - \int_{0}^{x} (-\cos t) \, dt \]

### Step 3: Evaluate the Boundary Term
Calculating the boundary term:
\[ \left[ t (-\cos t) \right]_{0}^{x} = -x \cos x - (0 \cdot (-\cos 0)) = -x \cos x \]

### Step 4: Evaluate the Remaining Integral
Now we need to evaluate the integral:
\[ \int_{0}^{x} \cos t \, dt \]
The integral of \( \cos t \) is \( \sin t \), so:
\[ \int_{0}^{x} \cos t \, dt = \left[ \sin t \right]_{0}^{x} = \sin x - \sin 0 = \sin x \]

### Step 5: Combine Results
Putting everything together, we have:
\[ f(x) = -x \cos x + \sin x \]

### Step 6: Differentiate \( f(x) \)
Now, we differentiate \( f(x) \) to find \( f'(x) \):
\[ f'(x) = \frac{d}{dx}(-x \cos x + \sin x) \]
Using the product rule on the first term:
\[ f'(x) = -\left( \cos x + x (-\sin x) \right) + \cos x \]
This simplifies to:
\[ f'(x) = -\cos x + x \sin x + \cos x = x \sin x \]

Thus, the final answer is:
\[ f'(x) = x \sin x \]