Evaluate: `int_(0)^(1//2) (sin^(-1)x)/((1-x^(2))^(3//2)) dx`

Answer

To evaluate the integral

\[ I = \int_{0}^{\frac{1}{2}} \frac{\sin^{-1} x}{(1 - x^2)^{\frac{3}{2}}} \, dx, \]

we will use a substitution method. Let's follow the steps:

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

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

### Step 2: Change the limits
When \( x = 0 \), \( t = \sin^{-1}(0) = 0 \).

When \( x = \frac{1}{2} \), \( t = \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \).

Thus, the limits change from \( 0 \) to \( \frac{\pi}{6} \).

### Step 3: Rewrite the integral
Substituting \( x \) and \( dx \) into the integral, we get:

\[ I = \int_{0}^{\frac{\pi}{6}} \frac{t}{(1 - \sin^2 t)^{\frac{3}{2}}} \cos t \, dt. \]

Using the identity \( 1 - \sin^2 t = \cos^2 t \), we have:

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

Therefore, the integral simplifies to:

\[ I = \int_{0}^{\frac{\pi}{6}} \frac{t \cos t}{\cos^3 t} \, dt = \int_{0}^{\frac{\pi}{6}} \frac{t}{\cos^2 t} \, dt. \]

### Step 4: Rewrite the integral
We can rewrite \( \frac{1}{\cos^2 t} \) as \( \sec^2 t \):

\[ I = \int_{0}^{\frac{\pi}{6}} t \sec^2 t \, dt. \]

### Step 5: Integration by parts
Let \( u = t \) and \( dv = \sec^2 t \, dt \). Then, we have:

\[ du = dt \quad \text{and} \quad v = \tan t. \]

Using integration by parts:

\[ I = \left[ t \tan t \right]_{0}^{\frac{\pi}{6}} - \int_{0}^{\frac{\pi}{6}} \tan t \, dt. \]

### Step 6: Evaluate the first term
Evaluating the first term:

\[ \left[ t \tan t \right]_{0}^{\frac{\pi}{6}} = \left( \frac{\pi}{6} \tan\left(\frac{\pi}{6}\right) \right) - (0 \cdot \tan(0)) = \frac{\pi}{6} \cdot \frac{1}{\sqrt{3}} = \frac{\pi}{6\sqrt{3}}. \]

### Step 7: Evaluate the integral
Now we need to evaluate \( \int_{0}^{\frac{\pi}{6}} \tan t \, dt \):

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

Thus,

\[ \int_{0}^{\frac{\pi}{6}} \tan t \, dt = \left[-\log|\cos t|\right]_{0}^{\frac{\pi}{6}} = -\log\left(\cos\left(\frac{\pi}{6}\right)\right) + \log\left(\cos(0)\right). \]

Calculating this gives:

\[ -\log\left(\frac{\sqrt{3}}{2}\right) + \log(1) = -\log\left(\frac{\sqrt{3}}{2}\right) = \log\left(\frac{2}{\sqrt{3}}\right). \]

### Step 8: Combine results
Putting it all together:

\[ I = \frac{\pi}{6\sqrt{3}} - \log\left(\frac{2}{\sqrt{3}}\right). \]

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

\[ I = \frac{\pi}{6\sqrt{3}} - \log\left(\frac{2}{\sqrt{3}}\right). \]