Evaluate $ \int\sqrt{1-x^3}-\sqrt[3]{1-x^2}dx $ and $ \int_0^1\sqrt{1-x^3}-\sqrt[3]{1-x^2}dx $

Question

How can we solve the following indefinite integral $$ \int\sqrt{1-x^3}-\sqrt[3]{1-x^2}dx? $$

What about $$\int_0^1\sqrt{1-x^3}-\sqrt[3]{1-x^2}dx?$$

Answer 1

Each term is continuous on $[0,1]$, so we don't need to worry about convergence. One way to see the definite integral is zero is to write the terms separately and transform one into the other: $$ \int _0^1 \sqrt{1-x^3}\,dx $$ $$ \stackrel{x=t^{1/3}}{\Rightarrow} \frac{1}{3} \int_0^1 t^{-2/3}(1-t)^{1/2} \,dt$$ $$ \stackrel{IBP}{=}\left.t^{1/3}(1-t)^{1/2}\right|_0^1+\frac{1}{2}\int_0^1 t^{1/3}(1-t)^{-1/2}\,dt $$ $$ =0+\frac{1}{2}\int_0^1 t^{1/3}(1-t)^{-1/2}\,dt $$ $$ \stackrel{x=\sqrt{1-t}}{\Rightarrow}\int _0^1 \sqrt[3]{1-x^2}\,dx $$Another approach is to observe that each term is a special case of the Beta function $B(a,b)$, namely for $0<\Re(a),\Re(b)$: $$ \int _0^1 x^{a-1}(1-x)^{b-1}\,dx = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}=:B(a,b) $$Post substitution, we have $$ \frac{1}{3} \int_0^1 t^{-2/3}(1-t)^{1/2} = \frac{\Gamma(1/3)\Gamma(3/2)}{3\Gamma(11/6)} = \frac{\Gamma(4/3)\Gamma(1/2)}{2\Gamma(11/6)} = \frac{1}{2}\int_0^1 t^{1/3}(1-t)^{-1/2}\,dt, $$again showing their difference is zero.

A closed-form for the indefinite integral is probably hopeless. The best one could do is a power-series or hypergeometric function; for instance, $$ \int \sqrt{1-x^3}\,dx = \int 1- \sum_{n=1}^{\infty}\frac{\left(n-\frac{3}{2}\right)!}{2 \sqrt{\pi } n!}x^{3n}\,dx = C+x- \sum_{n=1}^{\infty}\frac{\left(n-\frac{3}{2}\right)!}{2 \sqrt{\pi } n!(3n+1)}x^{3n+1} $$

Answer 2

Hint:

In given problem, notice that the two functions whose difference is given as the integrand are inverses of each other. And the negative sign comes about from reversing the limits from $[1,0]$ to $[0,1]$.

$$\int_{a}^{b}f(x)\mathrm dx+\int_{f(a)}^{f(b)}f^{-1}(x)\mathrm dx=bf(b)-af(a)$$