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

Explanation

To solve the integral \( I = \int \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} \, dx \), we will follow these steps:

### Step 1: Substitution
Let \( t = \sin^{-1} x \). Then, we have:
\[ x = \sin t \]
Differentiating both sides gives:
\[ dx = \cos t \, dt \]

### Step 2: Rewrite the Integral
Substituting \( x \) and \( dx \) into the integral, we get:
\[ I = \int \frac{t}{(1 - \sin^2 t)^{3/2}} \cos t \, dt \]
Using the identity \( 1 - \sin^2 t = \cos^2 t \), we can simplify the integral:
\[ I = \int \frac{t}{(\cos^2 t)^{3/2}} \cos t \, dt = \int \frac{t}{\cos^3 t} \cos t \, dt = \int \frac{t}{\cos^2 t} \, dt \]

### Step 3: Simplifying the Integral
Now we can rewrite the integral as:
\[ I = \int t \sec^2 t \, dt \]

### Step 4: Integration by Parts
Using integration by parts, let:
- \( u = t \) → \( du = dt \)
- \( dv = \sec^2 t \, dt \) → \( v = \tan t \)

Applying the integration by parts formula \( \int u \, dv = uv - \int v \, du \):
\[ I = t \tan t - \int \tan t \, dt \]

### Step 5: Integrate \( \tan t \)
The integral of \( \tan t \) is:
\[ \int \tan t \, dt = -\log |\cos t| + C \]
Thus, we have:
\[ I = t \tan t + \log |\cos t| + C \]

### Step 6: Substitute Back
Now substitute back \( t = \sin^{-1} x \):
\[ I = \sin^{-1} x \tan(\sin^{-1} x) + \log |\cos(\sin^{-1} x)| + C \]
Using the identity \( \tan(\sin^{-1} x) = \frac{x}{\sqrt{1 - x^2}} \) and \( \cos(\sin^{-1} x) = \sqrt{1 - x^2} \), we can write:
\[ I = \sin^{-1} x \cdot \frac{x}{\sqrt{1 - x^2}} + \log |\sqrt{1 - x^2}| + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} + \frac{1}{2} \log(1 - x^2) + C \]