Definite integral involving a square and a cube root I have the following integral to solve
$$\int_0^1 \left(\sqrt{1-x^3}-\sqrt[3]{1-x^2}\right) \, dx.$$
I tried the substitutions $u=x^2$, $u=x^3$, $u=\sqrt{1-x^3}$, and $u=\sqrt[3]{1-x^2}$ but the integral would net get any simpler. Any hint, how to tackle the integral is welcome (maybe I should try integration by parts?).
 A: Here is an informal argument:
First note that your integral is of the form $\int_0^1 f(x)-f^{-1}(x)\,dx$, where $f$ is a  decreasing function with $f(0)=1$, $f(1)=0$, and $f^{-1}$ is the inverse of $f$.
Then


*

*$ \int_0^1 f(x)\,dx$ is the area of the region bounded above by the graph of $f$
over the interval $0\le x\le 1$;


and, noting that the graph of the equation $x=f^{-1}(y)$ is precisely the graph of the equation $y=f(x)$,


*

*$ \int_0^1 f^{-1}(y)\,dy$ is the area of the region bounded to the right by the
graph of $f$, below by the interval $0\le x\le 1$, and to the left by
the interval $0\le y\le1$.


The two aforementioned regions coincide; thus, we have $\int_0^1 f(x)\,dx =\int_0^1 f^{-1}(x)\,dx $.
And so,  $\int_0^1 f(x)-f^{-1}(x)\,dx=0$.
A: You can use two substitutions and one integration by parts to prove that
$$\begin{equation*}
I=\int_{0}^{1}\sqrt[3]{1-x^{2}}dx=\int_{0}^{1}\sqrt{1-x^{3}}dx=J.
\end{equation*}$$
Starting with $I$ make the substitution $t=1-x^{2}$ to obtain
$$\begin{equation*}
I=\int_{0}^{1}\frac{\sqrt[3]{t}}{2\sqrt{1-t}}dt.
\end{equation*}$$
Now integrate by parts choosing the factors $u(t)=\sqrt[3]{t}$ and $v'(t)=\frac{1}{2\sqrt{1-t}}$ 
$$\begin{eqnarray*}
I &=&\left. \sqrt[3]{t}\left( -\sqrt{1-t}\right) \right\vert _{0}^{1}-\int_{0}^{1}\frac{1}{3\sqrt[3]{t^{2}}}\left( -\sqrt{1-t}\right) \,dt =\int_{0}^{1}\frac{1}{3\sqrt[3]{t^{2}}}\sqrt{1-t}\,dt.
\end{eqnarray*}$$
Finally use the substitution $t=v^{3}$
$$\begin{equation*}
I=\int_{0}^{1}\sqrt{1-v^{3}}\,dv=J.
\end{equation*}$$
