`intsinsqrt(x)dx=?`

Options

A. `-sqrt(x)cossqrt(x)+C`
B. `-sqrt(x)cossqrt(x)-2sinsqrt(x)-C`
C. `-2sqrt(x)cossqrt(x)+2sinsqrt(x)+C`
D. none of these

Explanation

To solve the integral \( \int \sin(\sqrt{x}) \, dx \), we can use a substitution method. Here’s the step-by-step solution:

### Step 1: Substitution
Let \( t = \sqrt{x} \). Then, we have:
\[ x = t^2 \]
Now, we differentiate both sides with respect to \( x \):
\[ dx = 2t \, dt \]

### Step 2: Rewrite the Integral
Now, we can rewrite the integral in terms of \( t \):
\[ \int \sin(\sqrt{x}) \, dx = \int \sin(t) \cdot 2t \, dt \]
This simplifies to:
\[ 2 \int t \sin(t) \, dt \]

### Step 3: Integration by Parts
We will use integration by parts to solve \( \int t \sin(t) \, dt \). Let:
- \( u = t \) (thus, \( du = dt \))
- \( dv = \sin(t) \, dt \) (thus, \( v = -\cos(t) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \), we have:
\[ \int t \sin(t) \, dt = -t \cos(t) - \int -\cos(t) \, dt \]
This simplifies to:
\[ -t \cos(t) + \int \cos(t) \, dt \]
The integral of \( \cos(t) \) is \( \sin(t) \), so we have:
\[ -t \cos(t) + \sin(t) \]

### Step 4: Substitute Back
Now, substituting back for \( t \):
\[ \int t \sin(t) \, dt = -\sqrt{x} \cos(\sqrt{x}) + \sin(\sqrt{x}) + C \]
Thus, multiplying by 2 (from Step 2):
\[ 2 \int t \sin(t) \, dt = -2\sqrt{x} \cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C \]

### Final Answer
The final result for the integral \( \int \sin(\sqrt{x}) \, dx \) is:
\[ \int \sin(\sqrt{x}) \, dx = -2\sqrt{x} \cos(\sqrt{x}) + 2\sin(\sqrt{x}) + C \]