TITLE: Definite integral involving a square and a cube root
QUESTION [5 upvotes]: 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?).

REPLY [4 votes]: 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 $t=v^{3}$
$$\begin{equation*}
I=\int_{0}^{1}\sqrt{1-v^{3}}\,dv=J.
\end{equation*}$$