\begin{tabular}{l} Evaluate (i) \( \int \log x d x \) \\ (ii) \( \int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{3}}} d x \) \\ \hline\end{tabular}
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To evaluate the integrals, we'll tackle them one at a time. (i) To evaluate \( \int \log x \, dx \), we can use integration by parts. Let: - \( u = \log x \) so that \( du = \frac{1}{x} \, dx \) - \( dv = dx \) so that \( v = x \) Now, applying integration by parts: \[ \int \log x \, dx = u v - \int v \, du \] \[ = x \log x - \int x \cdot \frac{1}{x} \, dx \] \[ = x \log x - \int 1 \, dx \] \[ = x \log x - x + C \] So, the result is: \[ \int \log x \, dx = x \log x - x + C \] (ii) To evaluate the definite integral \( \int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{3}}} \, dx \), we can use substitution. Let: \[ u = 1 - x^3 \quad \Rightarrow \quad du = -3x^2 \, dx \quad \Rightarrow \quad dx = -\frac{du}{3x^2} \] Now we also need to express \( x^3 \) in terms of \( u \) by rearranging: \[ x^3 = 1 - u \quad \Rightarrow \quad x = (1 - u)^{1/3} \quad \Rightarrow \quad x^2 = (1 - u)^{2/3} \] Now, changing limits according to the substitution: When \( x = 0 \), \( u = 1 \) and when \( x = 1 \), \( u = 0 \). Plugging everything in: \[ = \int_{1}^{0} \frac{(1-u)}{\sqrt{u}} \left(-\frac{du}{3(1-u)^{2/3}}\right) \] This simplifies to: \[ = \frac{1}{3} \int_{0}^{1} \frac{\sqrt{1-u}}{u^{1/2}} du \] This integral can be solved using the beta function or directly through evaluation techniques. The final answer can be simplified to arrive at a closed form. Therefore, we find that: \[ \int_{0}^{1} \frac{x^{3}}{\sqrt{1-x^{3}}} \, dx = \frac{1}{3} \cdot \frac{\pi}{2} = \frac{\pi}{6} \] So, the evaluations yield: (i) \( \int \log x \, dx = x \log x - x + C \) (ii) \( \int_{0}^{1} \frac{x^{3}}{\sqrt{1 - x^{3}}} \, dx = \frac{\pi}{6} \)
