# Tricky Integral involving radicals

I am trying to evaluate the following definite integral (for $a>0$):

$$I=\int_{0}^{1}{{{\left( {{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x \right)}^{2}}dx}$$ Neither the substitution $u={{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}$ nor $u={{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x$ are appropriate.I have also tried Feynman’s trick (differentiated with respect to a) but I didn’t get any success. Thanks in advance.

• How about $u=x^a$? – Chris2018 Jul 16 '18 at 15:06
• I am not sure let me try it – user577488 Jul 16 '18 at 15:09

Using the substitution $u={{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}$ $$I=\int_{0}^{1}{{{\left( {{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x \right)}^{2}}dx}=\int_{0}^{1}{{{u}^{a-1}}{{\left( 1-{{u}^{a}} \right)}^{\frac{1}{a}-1}}{{\left( u-{{\left( 1-{{u}^{a}} \right)}^{\frac{1}{a}}} \right)}^{2}}du}$$ Since both $x\ and\ u$ are dummy variables \begin{align} & I=\frac{1}{2}\int_{0}^{1}{{{\left( {{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x \right)}^{2}}+{{x}^{a-1}}{{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}-1}}{{\left( x-{{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}} \right)}^{2}}dx} \\ & \quad =\frac{1}{2}\int_{0}^{1}{{{\left( {{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x \right)}^{2}}\left( 1+{{x}^{a-1}}{{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}-1}} \right)dx} \\ & \quad =\frac{1}{2}\int_{0}^{1}{-\frac{1}{3}\frac{d}{dx}{{\left( {{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x \right)}^{3}}dx} \\ & \quad =\frac{1}{3} \\ \end{align}
Hint. By the change of variable $$u=x^a,\qquad x=u^{1/a},\qquad dx=\frac{1}{a}u^{1/a-1}du,$$ one gets $$I=\int_{0}^{1}{{{\left( {{\left( 1-{{x}^{a}} \right)}^{\frac{1}{a}}}-x \right)}^{2}}dx}=\frac{1}{a}\int_{0}^{1}{{{\left( {{\left( 1-u\right)}^{\frac{1}{a}}}-u^{1/a} \right)}^{2}}u^{1/a-1}du}$$ then by expanding the square one is led to apply the standard Euler beta evaluation: $$\int_{0}^{1}(1-u)^{s-1} u^{t-1}\,du = \frac{\Gamma(s)\Gamma(t)}{\Gamma(s+t)},\quad \operatorname{Re}(s)>0,\,\operatorname{Re}(t)>0.$$