91Ó°ÊÓ

In this case, we can choose the substitution: $$u = x^2$$ This will simplify the integrand and make it easier to see the integration process. We also need to calculate the corresponding derivative of our substitution: $$\frac{du}{dx} = 2x$$

Since our substitution is \(u = x^2\), we need to change the limits of integration accordingly: $$u(0) = 0^2 = 0$$ $$u\left(\frac{1}{2}\right) = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$$ So our new limits of integration will be \(u = 0\) and \(u = \frac{1}{4}\).

Taking into account our substitution \(u = x^2\) and \(\frac{du}{dx} = 2x\), we can replace the integrand and the differential as: $$\frac{x^2}{\sqrt{1 - x^2}} dx = \frac{u}{\sqrt{1 - u}} \frac{du}{2 \sqrt{u}}$$ Simplifying, we have: $$\frac{u}{\sqrt{1 - u}} \frac{du}{2 \sqrt{u}} = \frac{1}{2} \frac{u}{\sqrt{u(1 - u)}} du$$

Now we just need to evaluate the integral with our new limits of integration and simplified integrand: $$\int_{0}^{\frac{1}{4}} \frac{1}{2} \frac{u}{\sqrt{u(1 - u)}} du$$ To evaluate this integral, use the formula for integration by substitution: $$\frac{1}{2} \int_{0}^{\frac{1}{4}} \frac{1}{\sqrt{1 - u}} du$$ Now, using a standard integral table or software, we find that: $$\int \frac{1}{\sqrt{1 - u}} du = \arcsin(u) + C$$ So, $$\frac{1}{2} \int_{0}^{\frac{1}{4}} \frac{1}{\sqrt{1 - u}} du = \frac{1}{2} \left[\arcsin(u) \right]_{0}^{\frac{1}{4}}$$ By evaluating our expression at the limits of integration, we have: $$\frac{1}{2} \left[ \arcsin\left(\frac{1}{4}\right) - \arcsin(0) \right] = \frac{1}{2} \arcsin\left(\frac{1}{4}\right)$$ Thus, our solution is: $$\int_{0}^{\frac{1}{2}} \frac{x^2}{\sqrt{1 - x^2}} dx = \frac{1}{2} \arcsin\left(\frac{1}{4}\right)$$