Calculate the integral$$\int\limits_{0}^{3}\frac{\sqrt{\arcsin \frac{x}{3}}}{\sqrt[4]{9-x^2}}dx$$
I was only able to get a grade, but it gives me nothing. I can't think of a way to solve this integral. You can't use numerical methods if they don't give you an approximately accurate answer through non elementary functions. I was able to get a numerical approximation on the computer, but how do I solve the integral analytically?
$$\int\limits_{0}^{3}\frac{\sqrt{\arcsin \frac{x}{3}}}{\sqrt[4]{9-x^2}}dx \approx \sum_{i=0}^{n-1} \frac{h}{6} \left(\frac{\sqrt{\arcsin \frac{x_i}{3}}}{\sqrt[4]{9-x_i^2}} + 4\frac{\sqrt{\arcsin \frac{x_i + x_{i+1}}{6}}}{\sqrt[4]{9-(x_i + x_{i+1})^2}} + \frac{\sqrt{\arcsin \frac{x_{i+1}}{3}}}{\sqrt[4]{9-x_{i+1}^2}}\right)$$
where $h = (3-0)/n$, $x_i = 0 + i \cdot h$, and $n$ is the number of sub-sections into which the interval $[0, 3]$ is divided
Best Answer
For the fun of using my favored $1,400^+$ years ond approximation of the cosine function $$\int_0^{\frac \pi 2}\sqrt{t \cos (t)}\,dt \sim \int_0^{\frac \pi 2} \sqrt{t\,\frac{\pi ^2-4 t^2}{\pi^2+t ^2}}\,dt$$ There is an antiderivative $$\int \sqrt{t\,\frac{\pi ^2-4 t^2}{\pi^2+t ^2}}\,dt=\frac{2}{3} t^{3/2} F_1\left(\frac{3}{4};-\frac{1}{2},\frac{1}{2};\frac{7}{4};\frac{4 t^2}{\pi ^2},-\frac{t^2}{\pi ^2}\right)$$ where appears the Appell hypergeometric function of two variables.
For the definite integral $$\int_0^{\frac \pi 2} \sqrt{t\,\frac{\pi ^2-4 t^2}{\pi^2+t ^2}}\,dt=\frac{\pi ^2 \, \Gamma \left(\frac{7}{4}\right)}{6\, \sqrt{2}\, \Gamma \left(\frac{9}{4}\right)}\, _2F_1\left(\frac{1}{2},\frac{3}{4};\frac{9}{4};-\frac{1}{4}\right)$$ which is $0.907705$ to be compared to the "exact" $0.907288$ obtained by numerical integration.
This can be justified by the fact that $$\int_0^{\frac \pi 2}\left(\sqrt{t \cos (t)}-\sqrt{t\,\frac{ \pi ^2-4 t^2}{\pi ^2+t^2}}\right)^2\,dt=1.594\times 10^{-6}$$