To solve the integral \( I = \int \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} \, dx \), we will use the substitution method and integration by parts. Let's go through the solution step by step.
### Step 1: Substitution
Let \( x = \sin \theta \). Then, we have:
\[
dx = \cos \theta \, d\theta
\]
Also, we know that:
\[
\sin^{-1} x = \theta
\]
And, since \( x = \sin \theta \):
\[
1 - x^2 = 1 - \sin^2 \theta = \cos^2 \theta
\]
Thus,
\[
(1 - x^2)^{3/2} = (\cos^2 \theta)^{3/2} = \cos^3 \theta
\]
### Step 2: Rewrite the Integral
Substituting these into the integral, we get:
\[
I = \int \frac{\theta}{\cos^3 \theta} \cos \theta \, d\theta = \int \frac{\theta}{\cos^2 \theta} \, d\theta = \int \theta \sec^2 \theta \, d\theta
\]
### Step 3: Integration by Parts
Now, we will use integration by parts. Let:
- \( u = \theta \) \(\Rightarrow du = d\theta\)
- \( dv = \sec^2 \theta \, d\theta \) \(\Rightarrow v = \tan \theta\)
Using the integration by parts formula:
\[
\int u \, dv = uv - \int v \, du
\]
we have:
\[
I = \theta \tan \theta - \int \tan \theta \, d\theta
\]
### Step 4: Integrate \( \tan \theta \)
The integral of \( \tan \theta \) is:
\[
\int \tan \theta \, d\theta = -\ln |\cos \theta| + C
\]
Thus, we can write:
\[
I = \theta \tan \theta + \ln |\cos \theta| + C
\]
### Step 5: Substitute Back
Now we substitute back \( \theta = \sin^{-1} x \) and \( \tan \theta = \frac{x}{\sqrt{1 - x^2}} \):
\[
I = \sin^{-1} x \cdot \frac{x}{\sqrt{1 - x^2}} + \ln |\cos(\sin^{-1} x)| + C
\]
Since \( \cos(\sin^{-1} x) = \sqrt{1 - x^2} \), we have:
\[
I = \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} + \ln(\sqrt{1 - x^2}) + C
\]
This simplifies to:
\[
I = \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} + \frac{1}{2} \ln(1 - x^2) + C
\]
### Final Answer
Thus, the final result for the integral is:
\[
\int \frac{\sin^{-1} x}{(1 - x^2)^{3/2}} \, dx = \frac{x \sin^{-1} x}{\sqrt{1 - x^2}} + \frac{1}{2} \ln(1 - x^2) + C
\]
