Evaluate: `int cos^(-1)sqrt(x)dx`

Answer

To evaluate the integral \( \int \cos^{-1}(\sqrt{x}) \, dx \), we will follow a systematic approach using substitution and integration by parts.

### Step-by-Step Solution:

1. **Substitution**:
Let \( \sqrt{x} = \cos(\theta) \). Then, squaring both sides gives:
\[ x = \cos^2(\theta) \]
Differentiating both sides with respect to \( \theta \):
\[ dx = -2 \cos(\theta) \sin(\theta) \, d\theta = -\sin(2\theta) \, d\theta \]

2. **Rewrite the Integral**:
Substitute \( \sqrt{x} \) and \( dx \) into the integral:
\[ \int \cos^{-1}(\sqrt{x}) \, dx = \int \cos^{-1}(\cos(\theta)) (-\sin(2\theta)) \, d\theta \]
Since \( \cos^{-1}(\cos(\theta)) = \theta \), we have:
\[ = -\int \theta \sin(2\theta) \, d\theta \]

3. **Integration by Parts**:
Let \( u = \theta \) and \( dv = \sin(2\theta) \, d\theta \). Then, we differentiate and integrate:
\[ du = d\theta, \quad v = -\frac{1}{2} \cos(2\theta) \]
Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
We get:
\[ -\int \theta \sin(2\theta) \, d\theta = -\left( -\frac{1}{2} \theta \cos(2\theta) - \int -\frac{1}{2} \cos(2\theta) \, d\theta \right) \]
Simplifying gives:
\[ = \frac{1}{2} \theta \cos(2\theta) + \frac{1}{4} \sin(2\theta) \]

4. **Substituting Back**:
Recall that \( \theta = \cos^{-1}(\sqrt{x}) \) and \( \cos(2\theta) = 2\cos^2(\theta) - 1 = 2x - 1 \), and \( \sin(2\theta) = 2\sin(\theta)\cos(\theta) = 2\sqrt{1-x} \cdot \sqrt{x} \):
\[ = \frac{1}{2} \cos^{-1}(\sqrt{x}) (2x - 1) + \frac{1}{4} (2\sqrt{x(1-x)}) \]

5. **Final Result**:
Therefore, the integral evaluates to:
\[ \int \cos^{-1}(\sqrt{x}) \, dx = \frac{1}{2} \cos^{-1}(\sqrt{x}) (2x - 1) - \frac{1}{2} \sqrt{x(1-x)} + C \]