(b) Using Assignment 13 Q1(a), prove that \[ \int_{-1}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} d x=\frac{\pi}{2} \]

To prove that \[ \int_{-1}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx = \frac{\pi}{2}, \] we can use symmetry and a substitution. ### Step 1: Symmetry of the Integral Notice that the integrand \(\frac{x^2}{\sqrt{1-x^2}}\) is an even function because: \[ \frac{(-x)^2}{\sqrt{1-(-x)^2}} = \frac{x^2}{\sqrt{1-x^2}}. \] Thus, we can simplify the integral: \[ \int_{-1}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx = 2 \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx. \] ### Step 2: Substitution Now, we will evaluate the integral \[ \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx. \] We can use the substitution \(x = \sin(\theta)\), which gives \(dx = \cos(\theta) \, d\theta\). The limits change as follows: - When \(x = 0\), \(\theta = 0\). - When \(x = 1\), \(\theta = \frac{\pi}{2}\). Now, substituting into the integral: \[ \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\theta)}{\sqrt{1-\sin^2(\theta)}} \cos(\theta) \, d\theta. \] Since \(\sqrt{1 - \sin^2(\theta)} = \cos(\theta)\), we have: \[ \int_{0}^{\frac{\pi}{2}} \frac{\sin^2(\theta)}{\cos(\theta)} \cos(\theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta. \] ### Step 3: Evaluating \(\int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta\) We can use the identity \(\sin^2(\theta) = \frac{1 - \cos(2\theta)}{2}\): \[ \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta = \int_{0}^{\frac{\pi}{2}} \frac{1 - \cos(2\theta)}{2} \, d\theta = \frac{1}{2} \int_{0}^{\frac{\pi}{2}} 1 \, d\theta - \frac{1}{2} \int_{0}^{\frac{\pi}{2}} \cos(2\theta) \, d\theta. \] Calculating these integrals: 1. \(\int_{0}^{\frac{\pi}{2}} 1 \, d\theta = \frac{\pi}{2}\). 2. \(\int_{0}^{\frac{\pi}{2}} \cos(2\theta) \, d\theta = \left[ \frac{\sin(2\theta)}{2} \right]_{0}^{\frac{\pi}{2}} = \frac{\sin(\pi)}{2} - \frac{\sin(0)}{2} = 0\). Thus, \[ \int_{0}^{\frac{\pi}{2}} \sin^2(\theta) \, d\theta = \frac{1}{2} \cdot \frac{\pi}{2} - \frac{1}{2} \cdot 0 = \frac{\pi}{4}. \] ### Step 4: Final Calculation Now we return to our earlier result: \[ \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx = \frac{\pi}{4}. \] Therefore, \[ \int_{-1}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx = 2 \int_{0}^{1} \frac{x^{2}}{\sqrt{1-x^{2}}} \, dx = 2 \cdot \frac{\pi}{4} = \frac{\pi}{2}. \] Thus, we have proven that