Question
Evaluate the following:
$\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$
$\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$
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Answer
Let $\text{I}=\int\frac{\sqrt{\text{x}}}{\sqrt{\text{a}^3-\text{x}^3}}\text{dx}$ $=\int\frac{\sqrt{\text{x}}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2-\Big(\text{x}^{\frac{3}{2}}\Big)^2}}$
Put $=\text{x}^{\frac{3}{2}}=\text{t}\Rightarrow\frac{3}{2}\text{x}^{\frac{1}{2}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{2}{3}\int\frac{\text{dt}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2}-\text{t}^2}$ $=\frac{2}{3}\sin^{-1}\frac{\text{t}}{\text{a}^{\frac{3}{2}}}+\text{C}$
$=\frac{2}{3}\sin^{-1}\frac{\text{x}^{\frac{3}{2}}}{\text{a}^{\frac{3}{2}}}+\text{C}$ $=\frac{2}{3}\sin^{-1}\sqrt{\frac{\text{x}^3}{\text{a}^3}}+\text{C}$
Put $=\text{x}^{\frac{3}{2}}=\text{t}\Rightarrow\frac{3}{2}\text{x}^{\frac{1}{2}}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\frac{2}{3}\int\frac{\text{dt}}{\sqrt{\Big(\text{a}^{\frac{3}{2}}\Big)^2}-\text{t}^2}$ $=\frac{2}{3}\sin^{-1}\frac{\text{t}}{\text{a}^{\frac{3}{2}}}+\text{C}$
$=\frac{2}{3}\sin^{-1}\frac{\text{x}^{\frac{3}{2}}}{\text{a}^{\frac{3}{2}}}+\text{C}$ $=\frac{2}{3}\sin^{-1}\sqrt{\frac{\text{x}^3}{\text{a}^3}}+\text{C}$
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