mysterious symmetry

An easier solution than @Karan Chatrath ' s solution would be:

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Note that the functions $f(x)=(1-x^{21})^{1/20}$ and $g(x)=(1-x^{20})^{1/21}$ are inverse functions.

There exists a very good property between a function and it's inverse i.e, in general, it holds true that $\int_{a}^{b} f(x) dx + \int_{f(a)}^{f(b)} f^{-1}(x) dx = bf(b)-af(a)$

This can be visualised by considering the area under the graph made by these functions and the coordinate axes.

Hence, in our case , $a=0 , b=1 , f(a)=1 , f(b)=0$ and hence the answer is simply $\boxed{0}$ .

Consider the integral:

$I(a,b) = \int_{0}^{1} \left(1-x^a\right)^{1/b} \ dx$

Taking $\blue{x^{a/2} = \sin{t}}$ and transforming the integral leads to the following. Intermediate simplifications and substitutions are left out:

$I(a,b) = \frac{2}{a} \int_{0}^{\pi/2} \left(\cos{t}\right)^{\frac{b+2}{b}} \left(\sin{t}\right)^{\frac{2-a}{a}} \ dt$

Recalling that the formula for the general integral:

$\int_{0}^{\pi/2} \left(\cos{t}\right)^m \left(\sin{t}\right)^n \ dt = \frac{\Gamma\left(\frac{m+1}{2}\right)\Gamma\left(\frac{n+1}{2}\right)}{2\Gamma\left(\frac{m+n+2}{2}\right)}$

Where: $\Gamma(x)$ is the Gamma function. Applying this formula for the considered integral results in:

$I(a,b) = \frac{\frac{1}{a}\Gamma\left(\frac{b+1}{b}\right)\Gamma\left(\frac{1}{a}\right)}{\Gamma\left(\frac{a+b+ab}{ab}\right)}$ $\implies \boxed{I(a,b) =\frac{\Gamma\left(\frac{b+1}{b}\right)\Gamma\left(\frac{1+a}{a}\right)}{\Gamma\left(\frac{a+b+ab}{ab}\right)}} \ \because \frac{1}{a}\Gamma\left(\frac{1}{a}\right)=\Gamma\left(\frac{1+a}{a}\right)$

The above step uses the property: $\red{n\Gamma(n) = \Gamma(n+1)}$ . This implies that: $I(a,b)=I(b,a)$ $\implies \int_{0}^{1} \left(1-x^{21}\right)^{1/20} \ dx = \int_{0}^{1} \left(1-x^{20}\right)^{1/21} \ dx$ $\implies \boxed{\int_{0}^{1} \left(\left(1-x^{21}\right)^{1/20}-\left(1-x^{20}\right)^{1/21}\right) \ dx=0}$

Similar solution as @Karan Chatrath 's

$\begin{aligned} I & = \int_0^1 \left(\sqrt[20]{1-x^{21}} - \sqrt[21]{1-x^{20}} \right) dx \\ & = \blue{\int_0^1 \sqrt[20]{1-x^{21}} \ dx} - \red{\int_0^1 \sqrt[21]{1-x^{20}} \ dx} & \small \blue{\text{Let }u = x^{21} \implies du = 21 x^{20} dx} \\ & = \blue{\int_0^1 \frac {u^{-\frac {20}{21}}(1-u)^\frac 1{20}}{21} du} - \red{\int_0^1 \frac {u^{-\frac {19}{20}}(1-v)^\frac 1{21}}{20} dv} & \small \red{\text{and }v = x^{20} \implies dv = 20 x^{19} dx} \\ & = \frac {B \left(\frac 1{21}, \frac {21}{20} \right)}{21} - \frac {B \left(\frac 1{20}, \frac {22}{21} \right)}{20} & \small \blue{\text{Beta function }B (m,n) = \frac {\Gamma(m) \Gamma(n)}{\Gamma(m+n)}} \\ & = \frac {\Gamma\left(\frac 1{21}\right) \blue{\Gamma \left(\frac {21}{20} \right)}}{21 \Gamma \left(\frac {461}{420}\right)} - \frac {\Gamma\left(\frac 1{20}\right) \blue{\Gamma \left(\frac {22}{21} \right)}}{20 \Gamma \left(\frac {461}{420}\right)} & \small \blue{\text{Gamma function }\Gamma (1+s) = s\Gamma (s)} \\ & = \frac {\Gamma\left(\frac 1{21}\right) \blue{\Gamma \left(\frac 1{20} \right)}}{21 \cdot \blue{20} \Gamma \left(\frac {461}{420}\right)} - \frac {\Gamma\left(\frac 1{20}\right) \blue{\Gamma \left(\frac 1{21} \right)}}{\blue{21} \cdot 20 \Gamma \left(\frac {461}{420}\right)} \\ & = \boxed 0 \end{aligned}$

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References:

• Beta function
• Gamma function