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

Answer

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

\[ \int u \, dv = uv - \int v \, du \]

### Step 1: Choose \( u \) and \( dv \)
Let:
- \( u = \sin^{-1} x \) (which is the inverse function)
- \( dv = dx \)

### Step 2: Differentiate \( u \) and integrate \( dv \)
Now, we need to find \( du \) and \( v \):
- Differentiate \( u \):
\[ du = \frac{1}{\sqrt{1 - x^2}} \, dx \]
- Integrate \( dv \):
\[ v = x \]

### Step 3: Apply the integration by parts formula
Substituting into the integration by parts formula:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x - \int x \cdot \frac{1}{\sqrt{1 - x^2}} \, dx \]

### Step 4: Simplify the remaining integral
Now, we need to evaluate the integral \( \int \frac{x}{\sqrt{1 - x^2}} \, dx \). We can use a substitution here.

Let:
\[ t = 1 - x^2 \implies dt = -2x \, dx \implies dx = -\frac{dt}{2x} \]

Also, we have:
\[ x = \sqrt{1 - t} \]

### Step 5: Substitute in the integral
Substituting these into the integral gives:
\[ \int \frac{x}{\sqrt{1 - x^2}} \, dx = \int \frac{\sqrt{1 - t}}{\sqrt{t}} \left(-\frac{dt}{2\sqrt{1 - t}}\right) \]
This simplifies to:
\[ -\frac{1}{2} \int \frac{1}{\sqrt{t}} \, dt \]

### Step 6: Evaluate the integral
The integral \( \int \frac{1}{\sqrt{t}} \, dt = 2\sqrt{t} + C \), so:
\[ -\frac{1}{2} \cdot 2\sqrt{t} = -\sqrt{t} \]

### Step 7: Substitute back for \( t \)
Recall that \( t = 1 - x^2 \):
\[ -\sqrt{1 - x^2} \]

### Step 8: Combine results
Now, substituting back into our expression from Step 3:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C \]

### Final Result
Thus, the integral evaluates to:
\[ \int \sin^{-1} x \, dx = x \sin^{-1} x + \sqrt{1 - x^2} + C \]

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