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

We can write it as $$I =  \displaystyle \int_{0}^{1}\sqrt[3]{1-x^7}dx-\int_{0}^{1}\sqrt[7]{1-x^3}dx$$
Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$
So  we get $$I =  \displaystyle \int_{0}^{1}\left(1-x^7\right)^{\frac{1}{3}}dx+\int_{1}^{0}\left(1-x^3\right)^{\frac{1}{7}}dx$$
Now Let $$\displaystyle f(x) = \left(1-x^{7}\right)^{\frac{1}{3}}\;,$$ Then $$f^{-1}(x) = (1-x^3)^{\frac{1}{7}}$$ and also $f(0) = 1$ and $f(1) =0$
So Integral $$\displaystyle I = \int_{0}^{1}f(x)dx+\int_{f(0)}^{f(1)}f^{-1}(x)dx$$
Now let $f^{-1}(x) = z\;,$ Then $x=f(z)$ So we get $dx = f'(z)dz$
So Integral $$\displaystyle I =\int_{0}^{1}f(x)dx+\int_{0}^{1}z\cdot f'(z)dz$$
Now Integration by parts for second Integral, We get
$$\displaystyle I  =\int_{0}^{1}f(x)dx+\left[z\cdot f(z)\right]_{0}^{1}-\int_{0}^{1}f(z)dz$$
So using $$\displaystyle \bullet\; \int_{a}^{b}f(z)dz = \int_{a}^{b}f(x)dx$$
So we get $$\displaystyle I =\int_{0}^{1}f(x)dx+f(1) -\int_{0}^{1}f(x)dx = f(1) =0$$
