If the value of the integral `int_(0)^(1//2) (x^(2))/((1-x^(2))^(3//2))` dx
is `(k)/(6)` then k is equal to :

Options

A. `2 sqrt(3) - pi `
B. `2 sqrt(3) + pi`
C. `3sqrt(2) + pi `
D. `3 sqrt(2) - pi `

Answer

To solve the integral

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

we will use the substitution \( x = \sin \theta \).

### Step 1: Substitution
Let \( x = \sin \theta \). Then, we have:

\[ dx = \cos \theta \, d\theta. \]

The limits change as follows:
- When \( x = 0 \), \( \theta = 0 \).
- When \( x = \frac{1}{2} \), \( \sin \theta = \frac{1}{2} \) implies \( \theta = \frac{\pi}{6} \).

### Step 2: Rewrite the integral
Substituting \( x = \sin \theta \) into the integral, we get:

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

Since \( 1 - \sin^2 \theta = \cos^2 \theta \), we have:

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

Thus, the integral simplifies to:

\[ I = \int_{0}^{\frac{\pi}{6}} \frac{\sin^2 \theta}{\cos^3 \theta} \cos \theta \, d\theta = \int_{0}^{\frac{\pi}{6}} \frac{\sin^2 \theta}{\cos^2 \theta} \, d\theta = \int_{0}^{\frac{\pi}{6}} \tan^2 \theta \, d\theta. \]

### Step 3: Integrate
The integral of \( \tan^2 \theta \) can be expressed as:

\[ \tan^2 \theta = \sec^2 \theta - 1. \]

Thus,

\[ I = \int_{0}^{\frac{\pi}{6}} \tan^2 \theta \, d\theta = \int_{0}^{\frac{\pi}{6}} \sec^2 \theta \, d\theta - \int_{0}^{\frac{\pi}{6}} 1 \, d\theta. \]

Calculating these integrals:

1. The integral of \( \sec^2 \theta \) is \( \tan \theta \):

\[ \int_{0}^{\frac{\pi}{6}} \sec^2 \theta \, d\theta = \tan \left( \frac{\pi}{6} \right) - \tan(0) = \frac{1}{\sqrt{3}} - 0 = \frac{1}{\sqrt{3}}. \]

2. The integral of \( 1 \) is simply:

\[ \int_{0}^{\frac{\pi}{6}} 1 \, d\theta = \frac{\pi}{6} - 0 = \frac{\pi}{6}. \]

Putting it all together:

\[ I = \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right). \]

### Step 4: Final expression
Now we have:

\[ I = \frac{1}{\sqrt{3}} - \frac{\pi}{6}. \]

### Step 5: Express in terms of \( k \)
We know that \( I = \frac{k}{6} \). Therefore, we can equate:

\[ \frac{k}{6} = \frac{1}{\sqrt{3}} - \frac{\pi}{6}. \]

Multiplying through by 6 gives:

\[ k = 6 \left( \frac{1}{\sqrt{3}} - \frac{\pi}{6} \right) = \frac{6}{\sqrt{3}} - \pi = 2\sqrt{3} - \pi. \]

### Conclusion
Thus, the value of \( k \) is:

\[ \boxed{2\sqrt{3} - \pi}. \]