`int_(0)^(pi^(2)//4) sin sqrt(x)dx` equals to

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the integral \( \int_{0}^{\frac{\pi^2}{4}} \sin(\sqrt{x}) \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, \( x = t^2 \) and \( dx = 2t \, dt \).

### Step 2: Change the limits of integration
When \( x = 0 \), \( t = \sqrt{0} = 0 \).
When \( x = \frac{\pi^2}{4} \), \( t = \sqrt{\frac{\pi^2}{4}} = \frac{\pi}{2} \).

### Step 3: Rewrite the integral
The integral now becomes:
\[ \int_{0}^{\frac{\pi}{2}} \sin(t) \cdot 2t \, dt = 2 \int_{0}^{\frac{\pi}{2}} t \sin(t) \, dt \]

### Step 4: Integration by parts
We will use integration by parts where we let:
- \( u = t \) (thus \( du = dt \))
- \( dv = \sin(t) \, dt \) (thus \( v = -\cos(t) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ \int t \sin(t) \, dt = -t \cos(t) - \int -\cos(t) \, dt \]
\[ = -t \cos(t) + \int \cos(t) \, dt \]
\[ = -t \cos(t) + \sin(t) \]

### Step 5: Evaluate the integral from 0 to \( \frac{\pi}{2} \)
Now we need to evaluate:
\[ 2 \left[ -t \cos(t) + \sin(t) \right]_{0}^{\frac{\pi}{2}} \]

Calculating at the upper limit \( t = \frac{\pi}{2} \):
\[ -t \cos(t) + \sin(t) = -\frac{\pi}{2} \cdot \cos\left(\frac{\pi}{2}\right) + \sin\left(\frac{\pi}{2}\right) = 0 + 1 = 1 \]

Calculating at the lower limit \( t = 0 \):
\[ -t \cos(t) + \sin(t) = -0 \cdot \cos(0) + \sin(0) = 0 + 0 = 0 \]

Thus, we have:
\[ 2 \left[ 1 - 0 \right] = 2 \]

### Final Answer
The value of the integral \( \int_{0}^{\frac{\pi^2}{4}} \sin(\sqrt{x}) \, dx \) is \( \boxed{2} \).

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