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

$$I_1=\int_{0}^{1} (1-x^7)^{\frac{1}{3}}dx$$
substitute $t = 1-x^7$, $\,\,dt = -7x^6dx$, and by extension $\,\,x = \sqrt[7]{1-t}$
$$=-\frac{1}{7}\int_{0}^{1} \frac{\sqrt[3]{t}}{(1-t)^\frac{6}{7}}dx$$
Now integrate by parts choosing the factors $u = \sqrt[3]{t}$ $\,\, u' = \frac{1}{3t^{2/3}}$$\,\,dv = \frac{1}{(1-t)^{6/7}}\,\,$ and $v = -7(1-t)^{1/7}$
$$ = -\frac{1}{7}(-\sqrt[3]{t}\sqrt[7]{1-t})\bigg|_0^1 -\frac{1}{7}\int_0^1 \frac{-7\sqrt[7]{1-t}}{3t^{2/3}}dt$$
$$ = -\frac{1}{7}\int_0^1 \frac{-7\sqrt[7]{1-t}}{3t^{2/3}}dt$$
Finally, subsitute $t = v^3$, $\,\, dt = 3v^2dv$ 
$$ = \int_0^1 \sqrt[7]{1-v^3}dv$$
$$ = I_2$$
So what we have is
$$\int_{0}^{1} (I_2 - I_2)$$
$$=\int_{0}^{1} 0$$
$$=0$$
