Ta có:
$\int\limits_0^{\frac{\pi}{3}}(\sin^3x+\cos^3x)dx$
$=\int\limits_0^{\frac{\pi}{3}}\sin^3xdx+\int\limits_0^{\frac{\pi}{3}}\cos^3xdx$
$=-\int\limits_0^{\frac{\pi}{3}}(1-\cos^2x)d(\cos x)+\int\limits_0^{\frac{\pi}{3}}(1-\sin^2x)d(\sin x)$
$=-\left(\cos x-\dfrac{\cos^3x}{3}\right)\left|\begin{array}{l}\frac{\pi}{3}\\0\end{array}\right.+\left(\sin x-\dfrac{\sin^3x}{3}\right)\left|\begin{array}{l}\frac{\pi}{3}\\0\end{array}\right.=\dfrac{5+9\sqrt3}{24}$