Evaluate: `int(xsin^(-1)x)/sqrt(1-x^(2))dx`

Answer

To evaluate the integral \(\int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx\), we can follow these steps:

### Step 1: Substitution
Let \( t = \sin^{-1} x \). Then, we differentiate both sides with respect to \( x \):
\[ dt = \frac{1}{\sqrt{1 - x^2}} \, dx \quad \Rightarrow \quad dx = \sqrt{1 - x^2} \, dt \]
Also, since \( x = \sin t \), we can substitute \( x \) in terms of \( t \).

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral gives:
\[ \int \frac{\sin t \cdot t}{\sqrt{1 - \sin^2 t}} \sqrt{1 - \sin^2 t} \, dt \]
Since \(\sqrt{1 - \sin^2 t} = \cos t\), the integral simplifies to:
\[ \int t \sin t \, dt \]

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

Using the integration by parts formula \(\int u \, dv = uv - \int v \, du\):
\[ \int t \sin t \, dt = -t \cos t - \int -\cos t \, dt \]
This simplifies to:
\[ -t \cos t + \int \cos t \, dt = -t \cos t + \sin t + C \]

### Step 4: Substitute Back
Now we substitute back \( t = \sin^{-1} x \) and \( \sin t = x \):
\[ -t \cos t = -\sin^{-1} x \cdot \cos(\sin^{-1} x) \]
To find \(\cos(\sin^{-1} x)\), we use the identity:
\[ \cos(\sin^{-1} x) = \sqrt{1 - x^2} \]
Thus, we have:
\[ -t \cos t = -\sin^{-1} x \cdot \sqrt{1 - x^2} \]
And since \(\sin t = x\):
\[ \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = -\sin^{-1} x \cdot \sqrt{1 - x^2} + x + C \]

### Final Answer
The final result is:
\[ \int \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} \, dx = x - \sin^{-1} x \cdot \sqrt{1 - x^2} + C \]