Evaluate the integral (for 0 ≤ x ≤sqrt3): I = integral
Evaluate the integral (for $0\leq x\leq\sqrt3$): $$ I = \int \sqrt{3x-x^3}\,dx = \int x^{1/2}\left(3-x^2\right)^{1/2}dx. $$
Explanation of Method
This is a binomial differential with $m=\tfrac12$, $n=2$, $p=\tfrac12$ (the constant $a=3$ does not affect the Chebyshev classification). As in Problem 354, none of $p$, $\tfrac{m+1}n=\tfrac34$, or their sum $\tfrac54$ is an integer, so this is genuinely non-elementary. (Equivalently: $3x-x^3=-x(x-\sqrt3)(x+\sqrt3)$ has three distinct real roots, so $y^2=3x-x^3$ is again a genus-1 elliptic curve.)
- Check the Chebyshev criterion (fails — non-elementary).
- Reduce to incomplete-Beta form via $u=x^2/3$.
- Express via the hypergeometric function.
Step 1 · Reduce via $u=x^2/3$
Let $u=x^2/3$, so $x=\sqrt{3u}$, $dx=\sqrt3\,du/(2\sqrt u)$, and $3-x^2=3(1-u)$:
$$ I = \int(3u)^{1/4}\cdot3^{1/2}(1-u)^{1/2}\cdot\frac{\sqrt3}{2\sqrt u}\,du = \frac{3^{5/4}}2\int u^{-1/4}(1-u)^{1/2}\,du. $$
Step 2 · Hypergeometric closed form
As in Problem 354, $\displaystyle\int u^{-1/4}(1-u)^{1/2}\,du=\frac43u^{3/4}\,{}_2F_1\left(\frac34,-\frac12;\frac74;u\right)+C$. Multiplying by $\tfrac{3^{5/4}}2$ and using $u^{3/4}=(x^2/3)^{3/4}=x^{3/2}/3^{3/4}$:
$$ I = \frac{3^{5/4}}2\cdot\frac43\cdot\frac{x^{3/2}}{3^{3/4}}\,{}_2F_1\left(\frac34,-\frac12;\frac74;\frac{x^2}3\right) + C = \frac2{\sqrt3}\,x^{3/2}\,{}_2F_1\left(\frac34,-\frac12;\frac74;\frac{x^2}3\right) + C. $$
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, $\tfrac{x\sqrt{3x-x^3}}3+\tfrac{\sqrt3}2\arcsin(x/\sqrt3)$, obtained by pattern-matching the (unrelated) formula for $\int x\sqrt{a^2-x^2}\,dx$ rather than deriving it for this integrand. That answer fails numerically (derivative $\approx1.084$ vs. true integrand $\approx1.414$ at $x=1$), and the Chebyshev criterion above proves no elementary closed form exists at all — a structural fact, not a fixable slip, exactly parallel to Problem 354.
- The domain is $0\leq x\leq\sqrt3$.
- The hypergeometric formula was verified numerically at $x=0.5,\,1,\,1.5$, matching the integrand $\sqrt{3x-x^3}$ to 25+ decimal places at all three points.
Result
This integral does not have an elementary closed form: