Evaluate `intcos sqrt(x)dx`

Explanation

To evaluate the integral \( \int \cos(\sqrt{x}) \, dx \), we will follow these steps:

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

### Step 2: Rewrite the Integral
Substituting \( t \) and \( dx \) into the integral, we get:
\[ \int \cos(\sqrt{x}) \, dx = \int \cos(t) \cdot 2t \, dt = 2 \int t \cos(t) \, dt \]

### Step 3: Integration by Parts
Now we will use integration by parts. Let:
- \( u = t \) (which implies \( du = dt \))
- \( dv = \cos(t) \, dt \) (which implies \( v = \sin(t) \))

Using the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ 2 \int t \cos(t) \, dt = 2 \left( t \sin(t) - \int \sin(t) \, dt \right) \]

### Step 4: Integrate \( \sin(t) \)
The integral of \( \sin(t) \) is:
\[ \int \sin(t) \, dt = -\cos(t) \]
Thus, we substitute back:
\[ 2 \left( t \sin(t) - (-\cos(t)) \right) = 2 \left( t \sin(t) + \cos(t) \right) \]

### Step 5: Substitute Back to Original Variable
Now we substitute back \( t = \sqrt{x} \):
\[ = 2 \left( \sqrt{x} \sin(\sqrt{x}) + \cos(\sqrt{x}) \right) + C \]

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

---