Q35 / 75 · VITEEE 2021 · Mathematics
Let $\int \frac{x^{1 / 2}}{\sqrt{1-x^3}} d x=\frac{2}{3} \operatorname{gof}(x)+C$, then
- A$\mathrm{f}(\mathrm{x})=\sqrt{\mathrm{x}}$
- Correct answerB$\mathrm{f}(\mathrm{x})=\mathrm{x}^{3 / 2}$ and $\mathrm{g}(\mathrm{x})=\sin ^{-1} \mathrm{x}$
- C$f(x)=x^{2 / 3}$
- DNone of these
Solution
Put $\mathrm{x}^{3 / 2}=\mathrm{t} \Rightarrow \frac{3}{2} \mathrm{x}^{1 / 2} \mathrm{dx}=\mathrm{dt}$
$\therefore$ integral is
$$
\int \frac{\frac{2}{3} \mathrm{dt}}{\sqrt{1-\mathrm{t}^2}}=\frac{2}{3} \sin ^{-1} \mathrm{t}+\mathrm{C}=\frac{2}{3} \sin ^{-1}\left(\mathrm{x}^{3 / 2}\right)+\mathrm{C}
$$
Hence, the correct answer is option: 'B'.