Integral involving cube root and seventh root Find the value of $$\int_{0}^{1} (1-x^7)^{\frac{1}{3}}-(1-x^3)^{\frac{1}{7}}\:dx$$
My Approach:
Let $$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$ and 
$$I_2=\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dx$$
For $I_1$ substitute $x^7=1-t^3$ so $dx=\frac{-3t^2}{7}(1-t^3)^{\frac{-6}{7}}\:dt$ Hence
$$I_1=\int_{1}^{0}\frac{-3t^3}{7}(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$
$$I_1=\frac{-3}{7}\int_{0}^{1}(1-t^3-1)(1-t^3)^{\frac{-6}{7}}\:dt$$ $\implies$ 
$$I_1=\frac{-3I_2}{7}+\frac{3A}{7}$$
where $A=\int_{0}^{1}(1-t^3)^{\frac{-6}{7}}dt$
Similarly using substitution $x^3=1-t^7$ for $I_2$ and proceeding as above we get
$$I_2=\frac{-7I_1}{3}+\frac{7B}{3}$$ where $B=\int_{0}^{1}(1-t^7)^{\frac{-2}{3}}dt$
But we need to find $I_1-I_2$ so got stuck up here

Hint. A straightforward approach is to use the Euler beta function
$$
B(a,b) :=  \int _0^1 x^{a-1}(1-x)^{b-1}dx =\frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)},\quad a>0, b>0.
$$ Then you get
$$
\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dt-\int_{0}^{1} (1-x^3)^{\frac{1}{7}}dt=0.
$$
