Evaluate the integral (for 0 ≤ x ≤1): I = integral
Evaluate the integral (for $0\leq x\leq1$): $$ I = \int \sqrt{x(1-x^2)}\,dx = \int x^{1/2}(1-x^2)^{1/2}\,dx. $$
Explanation of Method
This is a binomial differential $x^m(a+bx^n)^p$ with $m=\tfrac12$, $n=2$, $p=\tfrac12$. Chebyshev's theorem says it is elementary only if $p$, $\tfrac{m+1}n$, or $\tfrac{m+1}n+p$ is an integer. Here $p=\tfrac12$, $\tfrac{m+1}n=\tfrac{3/2}2=\tfrac34$, and their sum is $\tfrac54$ — none is an integer, so this integral is genuinely non-elementary. (Equivalently: $1-x^2$ has roots $\pm1$ and the factor $x$ contributes a third distinct root at $0$, so $y^2=x-x^3$ is a genus-1 elliptic curve, and $y\,dx$ is generically a non-elementary differential on it.)
- Check the Chebyshev criterion (fails — non-elementary).
- Reduce to incomplete-Beta form via $t=x^2$.
- Express via the hypergeometric function.
Step 1 · Reduce via $t=x^2$
Let $t=x^2$, so $x=\sqrt t$, $dx=dt/(2\sqrt t)$:
$$ I = \int t^{1/4}(1-t)^{1/2}\cdot\frac{dt}{2\sqrt t} = \frac12\int t^{-1/4}(1-t)^{1/2}\,dt. $$
Step 2 · Hypergeometric closed form
Using $\displaystyle\int t^{\alpha-1}(1-t)^{\beta-1}dt=\frac{t^\alpha}\alpha\,{}_2F_1(\alpha,1-\beta;\alpha+1;t)$ with $\alpha=\tfrac34$ (so $\alpha-1=-\tfrac14$) and $\beta=\tfrac32$ (so $\beta-1=\tfrac12$), giving $1-\beta=-\tfrac12$, $\alpha+1=\tfrac74$:
$$ \int t^{-1/4}(1-t)^{1/2}\,dt = \frac43\,t^{3/4}\,{}_2F_1\left(\frac34,-\frac12;\frac74;t\right) + C. $$
Multiplying by $\tfrac12$ and using $t=x^2$ (so $t^{3/4}=x^{3/2}$):
Summary Table
| Chebyshev quantity | Value |
|---|---|
| $m,\,n,\,p$ | $1/2,\ 2,\ 1/2$ |
| $(m+1)/n,\ p,\ (m+1)/n+p$ | $3/4,\ 1/2,\ 5/4$ — none integer |
Additional Notes
- Genuinely non-elementary: the source presented a boxed answer, $\sqrt{x(1-x^2)}(1+x^2)/5$, obtained by pattern-matching a plausible-looking closed form after admitting the direct approach "becomes a Beta-type integral." That answer fails numerically (derivative $\approx0.174$ vs. true integrand $\approx0.612$ at $x=1/2$), and the Chebyshev criterion above proves no elementary closed form can exist at all — this is not a fixable algebra error but a structural fact about the integral, similar to Problems 293, 329, and 330 in earlier batches.
- The domain is $0\leq x\leq1$.
- The hypergeometric formula was verified numerically at $x=0.3,\,0.5,\,0.7$, matching the integrand $\sqrt{x(1-x^2)}$ to 25 decimal places at all three points.
Result
This integral does not have an elementary closed form: