Integrating $\frac{\int_0^{\sqrt\pi} 3\sqrt\theta\sin(\theta^2)\,d\theta}{\sqrt\pi-0}$ I've come across this  question (2b.) when studying for AP Calculus BC.
$$\frac{\int_0^{\sqrt\pi} 3\sqrt\theta\sin(\theta^2)\,d\theta}{\sqrt\pi-0}$$
I did the following, but I am stuck on how to reduce it from the sin and the radical.
$$\frac{3}{\sqrt\pi}\int_0^{\sqrt\pi} \sqrt\theta\sin(\theta^2)\,d\theta$$
$$\frac{3}{2\sqrt\pi}\int_0^{\pi}\frac{\sin(u)}{\sqrt[4]u}\,du$$ $$u=\theta^2, du = 2\theta\,d\theta$$
I also tried a partial integration route (plus another sub) which left me with
$$-\frac{2}{\sqrt\pi}\int_0^\pi\sqrt[4]{x^3}\cos(x)\,dx$$
Same issue, that I can't simplify the radical and trig function together.
Any ideas? You're also welcome to look at the problem and see if there is another approach.

Btw, I don't see any elementary solutions on WolframAlpha, so I don't know if it's possible to solve by hand.
 A: As a quick preface, this is simply the best answer I can give. This is most likely not the most efficient method to achieve the answer shown, but this is just the way I did it. Note that if you put the expression in WolframAlpha, it will spit out the same hypergeometric function. Since you wanted a way to get the answer with a basic calculator or by hand, just sum up the first 20 terms or so, and you'll get a pretty solid approximation (the series converges rather quickly). I personally prefer the summation notation, so here is the derivation:
$$I=\frac{1}{\sqrt{\pi}}\int_0^\sqrt{\pi}3\sqrt{\theta}\sin\left(\theta^2\right)d\theta$$
$$u=\sqrt{\theta},\ du=\frac{d\theta}{2\sqrt{\theta}},\ d\theta=2\sqrt{\theta}du$$
$$=\frac{2}{\sqrt{\pi}}\int_{0}^{\sqrt[4]{\pi}}3u^{2}\sin\left(u^{4}\right)du$$
$$v=u^3,\ dv=3u^2du$$
$$=\frac{2}{\sqrt{\pi}}\int_{0}^{\pi^{\frac{3}{4}}}\sin\left(v\sqrt[3]{v}\right)dv$$
$$=\frac{2}{\sqrt{\pi}}\int_{0}^{\pi^{\frac{3}{4}}}\sum_{n\ge0}\frac{\left(-1\right)^{n}\left(v\sqrt[3]{v}\right)^{2n+1}}{\left(2n+1\right)!}dv$$
$$Using\ Fubini\text{'}s\ Theorem,\ Switch\ the\ Order\ of\ Summation$$
$$=\frac{2}{\sqrt{\pi}}\sum_{n\ge0}\frac{\left(-1\right)^{n}}{\left(2n+1\right)!}\int_{0}^{\pi^{\frac{3}{4}}}v^{\frac{8n+4}{3}}dv$$
$$=6\pi^{\frac{5}{4}}\sum_{n\ge0}\frac{\left(-1\right)^{n}\pi^{2n}}{\left(8n+7\right)\left(2n+1\right)!}$$
$$We\ can\ now\ stop\ if\ you\ want\ a\ Real\ Answer,$$
$$But\ I\ personally\ like\ this\ simplification\ more:$$
$$=\frac{6\pi\sqrt[4]{\pi}}{i\pi}\sum_{n\in\{\mathrm{odd}\}}\frac{\left(i\pi\right)^{n}}{\left(n+3\right)n!}$$
$$=-6i\sqrt[4]{\pi}\sum_{n\in\{\mathrm{odd}\}}\frac{\left(i\pi\right)^{n}}{\left(n+3\right)n!}$$
