Evaluate `intcos sqrt(x)dx`

Explanation

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

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

### Step 2: Rewrite the Integral
Now, substitute \( t \) and \( dx \) into the integral:
\[ \int \cos(\sqrt{x}) \, dx = \int \cos(t) \cdot 2t \, dt \]
This simplifies to:
\[ 2 \int t \cos(t) \, dt \]

### Step 3: Integration by Parts
We will use integration by parts, which states:
\[ \int u \, dv = uv - \int v \, du \]
Let:
- \( u = t \) (thus \( du = dt \))
- \( dv = \cos(t) \, dt \) (thus \( v = \sin(t) \))

Now, applying integration by parts:
\[ \int t \cos(t) \, dt = t \sin(t) - \int \sin(t) \, dt \]
The integral of \( \sin(t) \) is:
\[ \int \sin(t) \, dt = -\cos(t) \]
So, we have:
\[ \int t \cos(t) \, dt = t \sin(t) + \cos(t) \]

### Step 4: Substitute Back
Now, substitute this back into our integral:
\[ 2 \int t \cos(t) \, dt = 2(t \sin(t) + \cos(t)) \]
Substituting \( t = \sqrt{x} \):
\[ = 2(\sqrt{x} \sin(\sqrt{x}) + \cos(\sqrt{x})) \]

### Step 5: Final Answer
Thus, the final result of the integral is:
\[ \int \cos(\sqrt{x}) \, dx = 2\sqrt{x} \sin(\sqrt{x}) + 2\cos(\sqrt{x}) + C \]
where \( C \) is the constant of integration.

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