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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.

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    $\begingroup$ Try to first change variables $u = \sqrt{x}$ so that $I = \int_0^{\pi/2} 14 u \sin(u) du$. Now you can do integration by parts. $\endgroup$ – Winther Aug 27 '14 at 1:24
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    $\begingroup$ You should start with $u=\sqrt{x}$. Since $x=u^2$, $dx=2udu$, and then you have $2 \int u \sin(u) du$. If you were good at other problems, it should be clear how to proceed from there. The limits they gave you sort of suggest that this is what you should do, by the way. $\endgroup$ – Ian Aug 27 '14 at 1:25
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    $\begingroup$ @Kenshin He just did a $u$ substitution. With the integral $\int \limits_{0}^{\frac{\pi^{2}}{4}} 7 \sin{(\sqrt{x})} \,dx$, if you let $u = \sqrt{x}$, then we can square both sides to get $u^{2} = x$. Differentiating both sides gives $2udu = dx$. So now we can substitute into the original problem $u$ in place of $\sqrt{x}$ and $2udu$ in place of $dx$. That's why $\int \limits_{0}^{\frac{\pi^{2}}{4}} 7 \sin{(\sqrt{x})} \,dx =\int \limits_{0}^{\frac{\pi}{2}} (7)(2u) \sin{(u)} \,du $. $\endgroup$ – layman Aug 27 '14 at 1:55
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    $\begingroup$ @Kenshin Notice that the limits of integration change when you do $u$-substitution. The way to figure out the new limits is to look at the old limits. So with the substitution $u = \sqrt{x}$, we have when $x = 0$, $u = \sqrt{0} = 0$, so the new lower limit is $0$. When $x = \frac{\pi^{2}}{4}$, $u = \sqrt{\frac{\pi^{2}}{4}} = \frac{\pi}{2}$, so the new upper limit is $\frac{\pi}{2}$. $\endgroup$ – layman Aug 27 '14 at 1:57
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    $\begingroup$ @Kenshin Also, Ian's answer is incomplete. You don't get $2\int u \sin{u} \,du$, because we have the constant $7$ to deal with in the original problem. So it actually becomes $\int (7)(2) u \sin{u} \,du = \int 14 u \sin{u} \,du$. $\endgroup$ – layman Aug 27 '14 at 1:58
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$$I=\int_{0}^{\pi^2/4}7\sin(\sqrt{x})dx$$ Let $u=\sqrt x$, $x=u^2$, $dx=2udu$ $$I'=14\int u\sin(u)du=14\left(u\int \sin u du-\int \left(\frac d{du}u\right)\left(\int \sin u du\right)du\right) \\=14(-u\cos u+\sin u)$$ $$I=14(-u\cos u+\sin u)_0^{\pi/2}=14$$

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  • $\begingroup$ Great , thanks a lot! $\endgroup$ – Kenshin Aug 27 '14 at 14:36

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