Evaluate : `int(x^(2) sin^(-1)x)/((1-x^(2))^((3)/(2)))dx`

Answer

To evaluate the integral

\[ \int \frac{x^2 \sin^{-1} x}{(1 - x^2)^{3/2}} \, dx, \]

we can follow these steps:

### Step 1: Substitution
Let \( t = \sin^{-1} x \). Then, we have:

\[ x = \sin t \quad \text{and} \quad dx = \cos t \, dt. \]

Also, we know that:

\[ 1 - x^2 = 1 - \sin^2 t = \cos^2 t. \]

Thus,

\[ (1 - x^2)^{3/2} = (\cos^2 t)^{3/2} = \cos^3 t. \]

### Step 2: Rewrite the Integral
Substituting these into the integral gives:

\[ \int \frac{(\sin^2 t)(t)}{\cos^3 t} \cos t \, dt = \int \frac{t \sin^2 t}{\cos^2 t} \, dt = \int t \tan^2 t \, dt. \]

### Step 3: Integration by Parts
Now, we can use integration by parts. Let:

- \( u = t \) (then \( du = dt \))
- \( dv = \tan^2 t \, dt \) (then \( v = \tan t - t \))

Using integration by parts:

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

we have:

\[ \int t \tan^2 t \, dt = t \tan t - \int \tan t \, dt. \]

### Step 4: Integrate \( \tan t \)
The integral of \( \tan t \) is:

\[ \int \tan t \, dt = -\log |\cos t| + C. \]

### Step 5: Combine Results
Putting it all together, we get:

\[ \int t \tan^2 t \, dt = t \tan t + \log |\cos t| + C. \]

### Step 6: Back Substitute
Now we substitute back \( t = \sin^{-1} x \):

\[ = \sin^{-1} x \tan(\sin^{-1} x) + \log |\cos(\sin^{-1} x)| + C. \]

Since \( \tan(\sin^{-1} x) = \frac{x}{\sqrt{1 - x^2}} \) and \( \cos(\sin^{-1} x) = \sqrt{1 - x^2} \), we can write:

\[ = \sin^{-1} x \cdot \frac{x}{\sqrt{1 - x^2}} + \log(\sqrt{1 - x^2}) + C. \]

### Final Answer
Thus, the final answer is:

\[ \sin^{-1} x \cdot \frac{x}{\sqrt{1 - x^2}} + \frac{1}{2} \log(1 - x^2) + C. \]

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