Evaluate `int sin^(-1)x dx`.

Answer

To evaluate the integral \( \int \sin^{-1} x \, dx \), we will use the method of integration by parts.

### Step-by-step Solution:

1. **Identify \( u \) and \( dv \)**:
We choose:
\[ u = \sin^{-1} x \quad \text{and} \quad dv = dx \]

2. **Differentiate \( u \) and integrate \( dv \)**:
Now we find \( du \) and \( v \):
\[ du = \frac{1}{\sqrt{1 - x^2}} \, dx \quad \text{and} \quad v = x \]

3. **Apply the integration by parts formula**:
The integration by parts formula is given by:
\[ \int u \, dv = uv - \int v \, du \]
Substituting our values, we get:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx \]

4. **Simplify the remaining integral**:
We need to evaluate \( \int \frac{x}{\sqrt{1 - x^2}} \, dx \). To do this, we can use a substitution:
Let \( t = 1 - x^2 \). Then, differentiating gives:
\[ dt = -2x \, dx \quad \Rightarrow \quad dx = -\frac{dt}{2x} \]
Substituting \( x^2 = 1 - t \) into the integral:
\[ \int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{x}{\sqrt{t}} \left(-\frac{dt}{2x}\right) = -\frac{1}{2} \int t^{-1/2} \, dt \]

5. **Integrate**:
The integral \( \int t^{-1/2} \, dt \) is:
\[ -\frac{1}{2} \cdot 2 t^{1/2} = -t^{1/2} \]
Substituting back for \( t \):
\[ -\sqrt{1 - x^2} \]

6. **Combine results**:
Now substituting back into our integration by parts result:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C \]

### Final Answer:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C \]