`int(sin^(-1)x)/(sqrt(1-x^(2)))dx`

Answer

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

### Step 1: Substitution
Let \( t = \sin^{-1} x \). Then, we know that:
\[ x = \sin t \]
Differentiating both sides with respect to \( x \):
\[ \frac{dx}{dt} = \cos t \quad \text{or} \quad dx = \cos t \, dt \]
Also, we can express \( \sqrt{1 - x^2} \) in terms of \( t \):
\[ \sqrt{1 - x^2} = \sqrt{1 - \sin^2 t} = \cos t \]

### Step 2: Change of Variables
Now, substituting \( x \) and \( dx \) in the integral:
\[ \int \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \, dx = \int \frac{t}{\cos t} \cdot \cos t \, dt \]
This simplifies to:
\[ \int t \, dt \]

### Step 3: Integrate
Now, we can integrate \( t \):
\[ \int t \, dt = \frac{t^2}{2} + C \]

### Step 4: Back Substitution
Now, we substitute back \( t = \sin^{-1} x \):
\[ \frac{t^2}{2} + C = \frac{(\sin^{-1} x)^2}{2} + C \]

### Final Answer
Thus, the integral is:
\[ \int \frac{\sin^{-1} x}{\sqrt{1 - x^2}} \, dx = \frac{(\sin^{-1} x)^2}{2} + C \]

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