Find $\int \frac{ydy}{\sqrt{a^2-y^3}}$ $\int \frac{ydy}{\sqrt{a^2-y^3}}$ So I solved this using U-substituion where my  $u=a^2-y^3$ so, my $\frac{du}{-3y^2}$$=dy$ Making my answer $-$$\frac{2}{3x\sqrt{a^2-y^3}}$ I tried to use wolfram to check my answer but the answer in wolfram is way more different and complicated. So, my question is am I wrong? I think so, and I remember my professor saying that you can only use U-substitution when the difference of exponents is only 1. Is that true? Also kindly tell me how to solve problems like this 
 A: Assume $a>0$ and $y_0\le y_1\le a^{2/3}$.
Substitute $y=a^{2/3}x$, and define $x_{0,1}:=a^{-2/3}y_{0,1}$. Then $x_0\le x_1\le 1$, and:
$$\begin{align}
\int_{y_0}^{y_1}\frac{y\,\mathrm{d}y}{\sqrt{a^2-y^3}}
&=\int_{y_0a^{-2/3}}^{y_1a^{-2/3}}\frac{a^{2/3}x\cdot a^{2/3}\,\mathrm{d}x}{\sqrt{a^2-a^3x^3}}\\
&=\sqrt[3]{a}\int_{x_0}^{x_1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}}\\
&=\sqrt[3]{a}\int_{x_0}^{1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}}+\sqrt[3]{a}\int_{1}^{x_1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}}\\
&=\sqrt[3]{a}\int_{x_0}^{1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}}-\sqrt[3]{a}\int_{x_1}^{1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}}.\\
\end{align}$$
This reduces the problem to that of solving the definite integral,
$$\int_{u}^{1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}};~\text{where }u<1.\tag{1}$$
The definite integral $(1)$ vanishes identically when $u=1$ (trivially), and for $u=0$ we can express the integral as a beta function:
$$\begin{align}
\int_{0}^{1}\frac{x\,\mathrm{d}x}{\sqrt{1-x^3}}
&=\int_{0}^{1}\frac{t^{1/3}\cdot\frac13t^{-2/3}\,\mathrm{d}t}{\sqrt{1-t}}\\
&=\frac13\int_{0}^{1}\frac{t^{-1/3}\,\mathrm{d}t}{\sqrt{1-t}}\\
&=\frac13\operatorname{B}{\left(\frac23,\frac12\right)}.\\
\end{align}$$
Other than the special cases where $u=0$ or $u=1$ though, in general evaluating the definite integral $(1)$ will require elliptic integrals. If your textbook does not cover elliptic integrals, there is a good chance the question has a typo.
