`int_(0)^(1//2) (x sin^(-1)x)/(sqrt(1-x^(2)))dx=?`

Options

A. `-(pi sqrt(3))/(12)+(1)/(2)`
B. `(pisqrt(3))/(12)-(1)/(2)`
C. `(pisqrt(3))/(12) -(1)/(2)`
D. None of these

Answer

To solve the integral

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

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

### Step 1: Substitution
Let \( x = \sin(\theta) \). Then, we differentiate to find \( dx \):

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

### Step 2: Change of Limits
Now, we need to change the limits of integration. When \( x = 0 \):

\[ \sin(\theta) = 0 \implies \theta = 0. \]

When \( x = \frac{1}{2} \):

\[ \sin(\theta) = \frac{1}{2} \implies \theta = \frac{\pi}{6}. \]

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

### Step 3: Rewrite the Integral
Now substituting \( x \) and \( dx \) into the integral:

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

Since \( \sin^{-1}(\sin(\theta)) = \theta \) and \( \sqrt{1 - \sin^2(\theta)} = \cos(\theta) \), we can simplify the integral:

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

### Step 4: Simplifying the Integral
Now, we can express the integral as:

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

### Step 5: Integration by Parts
We will use integration by parts, where we let:

- \( u = \theta \) and \( dv = \sin(\theta) \cos(\theta) \, d\theta \).

Then, we differentiate and integrate:

- \( du = d\theta \)
- \( v = \frac{1}{2} \sin^2(\theta) \) (since \( \sin(\theta) \cos(\theta) = \frac{1}{2} \sin(2\theta) \))

Using integration by parts:

\[ I = \left[ \theta \cdot \frac{1}{2} \sin^2(\theta) \right]_{0}^{\frac{\pi}{6}} - \int_{0}^{\frac{\pi}{6}} \frac{1}{2} \sin^2(\theta) \, d\theta. \]

### Step 6: Evaluate the First Term
Calculating the first term:

\[ \left[ \theta \cdot \frac{1}{2} \sin^2(\theta) \right]_{0}^{\frac{\pi}{6}} = \frac{\pi}{6} \cdot \frac{1}{2} \left(\frac{1}{2}\right)^2 - 0 = \frac{\pi}{6} \cdot \frac{1}{8} = \frac{\pi}{48}. \]

### Step 7: Evaluate the Second Integral
Now, we need to evaluate:

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

Using the identity \( \sin^2(\theta) = \frac{1 - \cos(2\theta)}{2} \):

\[ \int_{0}^{\frac{\pi}{6}} \frac{1}{2} \sin^2(\theta) \, d\theta = \frac{1}{4} \int_{0}^{\frac{\pi}{6}} (1 - \cos(2\theta)) \, d\theta = \frac{1}{4} \left[ \theta - \frac{1}{2} \sin(2\theta) \right]_{0}^{\frac{\pi}{6}}. \]

Evaluating this gives:

\[ \frac{1}{4} \left[ \frac{\pi}{6} - \frac{1}{2} \sin\left(\frac{\pi}{3}\right) \right] = \frac{1}{4} \left[ \frac{\pi}{6} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2} \right] = \frac{1}{4} \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right]. \]

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

\[ I = \frac{\pi}{48} - \frac{1}{4} \left[ \frac{\pi}{6} - \frac{\sqrt{3}}{4} \right]. \]

This simplifies to:

\[ I = \frac{\pi}{48} - \frac{\pi}{24} + \frac{\sqrt{3}}{16}. \]

### Final Answer
After simplifying, we get:

\[ I = \frac{\sqrt{3}}{16} - \frac{\pi}{48}. \]