How do I integrate $\int_{0}^{\frac{\pi^2}{4}}7\sin(\sqrt{x})dx$?

Question

So, quick backstory. My semester just started and we are starting off by learning integration by parts. Which hasn't caused me much trouble except for this problem. $$\int_{0}^{\frac{\pi^2}{4}}7\sin(\sqrt{x})dx$$ I 'm going to post my steps that I have taken so far, but I stopped since I don't seem to be going anywhere. $$7\int_{0}^{\frac{\pi^2}{4}}\sin(\sqrt{x})dx u=\sin(\sqrt{x})$$ $u=\sin(\sqrt{x})$, $du=\frac{1}{2}\cos(\sqrt{x})*x^{\frac{-1}{2}}$, $dv=dx$,$v=x$ So applying $\int udv=uv-\int vdu$ I end with $$x\sin(\sqrt{x})-\int x\frac{1}{2}\cos(\sqrt{x})*x^{\frac{-1}{2}} \rightarrow x\sin(\sqrt{x})-\frac{1}{2}\int \cos(\sqrt{x})*x^{\frac{1}{2}} $$ (Left the constant 7 out until the end) Then I do the same to the remaining integral for 2 more times until i realized I'm not going anywhere. So a detailed procedure on how to solve this integral would be greatly appreciated, thanks in advance.

Answer 1

$$I=\int_{0}^{\pi^2/4}7\sin(\sqrt{x})dx$$ Let $u=\sqrt x$, $x=u^2$, $dx=2udu$ $$I'=14\int u\sin(u)du=14\left(u\int \sin u du-\int \left(\frac d{du}u\right)\left(\int \sin u du\right)du\right) \\=14(-u\cos u+\sin u)$$ $$I=14(-u\cos u+\sin u)_0^{\pi/2}=14$$