<s>[INST] 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
 [/INST] Consider the area in the first quadrant bounded by the curve $x^7+y^3 = 1$.

The curve in the first quadrant can be written as $y = (1-x^7)^{1/3}$ for $0 \le x \le 1$. Hence, the area bounded by the curve is given by $\displaystyle\int_{0}^{1}(1-x^7)^{1/3}\,dx$. 
We can also write the curve in the first quadrant as $x = (1-y^3)^{1/7}$ for $0 \le y \le 1$. So, the area bounded by the curve is also $\displaystyle\int_{0}^{1}(1-y^3)^{1/7}\,dy$.
Therefore, $\displaystyle\int_{0}^{1}(1-x^7)^{1/3}\,dx = \int_{0}^{1}(1-y^3)^{1/7}\,dy$, and thus, the difference is $\displaystyle\int_{0}^{1}\left[(1-x^7)^{1/3} - (1-x^3)^{1/7}\right]\,dx = 0 $.
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