Evaluation of $ \int_{0}^{1}\left(\sqrt[4]{1-x^7}-\sqrt[7]{1-x^4}\right)dx$ 
Evaluation of $\displaystyle \int_{0}^{1}\left(\sqrt[4]{1-x^7}-\sqrt[7]{1-x^4}\right)dx$

$\bf{My\; Try::}$ We can write it as $$I =  \displaystyle \int_{0}^{1}\sqrt[4]{1-x^7}dx-\int_{0}^{1}\sqrt[7]{1-x^4}dx$$
Now Using $$\displaystyle \bullet \int_{a}^{b}f(x)dx = -\int_{b}^{a}f(x)dx$$
So  we get $$I =  \displaystyle \int_{0}^{1}\left(1-x^7\right)^{\frac{1}{4}}dx+\int_{1}^{0}\left(1-x^4\right)^{\frac{1}{7}}dx$$
Now Let $$\displaystyle f(x) = \left(1-x^{7}\right)^{\frac{1}{4}}\;,$$ Then $$f^{-1}(x) = (1-x^4)^{\frac{1}{7}}$$ 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 $$\displaystyle I =\int_{0}^{1}f(x)dx+f(1) -\int_{0}^{1}f(x)dx = f(1) =0$$
My Question is can we solve it Some $\bf{short\;  way,}$ Iy yes then plz explain here
Thanks 

Hint: Notice that both integrands represent the same geometric shape, namely $X^4+Y^7=1$.
