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$$\int_0^a\int_0^1s\sqrt{q+s^2}\,dq\,ds$$ How do I solve this?

By substitution $u=q+s^2$ I get $$\frac23\int_0^a(1+s^2)^{\frac32}-s^4\,ds$$

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  • $\begingroup$ Add your computation, since there is a mistake in it. $\endgroup$ Sep 18, 2017 at 7:45

3 Answers 3

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hint: If $q > 0$, let $q = ks^2, k > 0$, and $q+s^2 = (k+1)s^2$,and you can take it from here. If $q < 0$, put $q = -c^2$ where $c > 0$, and let $s = c\sec \theta$. It is a bit tedious but surely doable...If $q = 0$, I am sure you can handle this....

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  • $\begingroup$ Since q is integrated in $(0,1)$ he probably only needs to consider the $q>0$ case. $\endgroup$
    – Sonal_sqrt
    Sep 18, 2017 at 7:48
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$$\newcommand{\i}[1]{\int_{0}^{#1}}$$ You went wrong here, $$\i{1}s(\sqrt{q+s^2})dq$$ $$s(1+s^2)^{\frac{3}{2}}-s^4$$

And for $$\i{a}\color{green}{s}\color{blue}{(1+s^2)^{3/2}}ds$$ Do it like this$$$$ Substituting $t^2=1+s^2$ $$ds=\frac{tdt}{\sqrt{1-t^2}}$$ $$\i{a} \color{blue}{t^3} \color{green}{\sqrt{1-t^2}} \frac{tdt}{\sqrt{1-t^2}}$$ $$\i{a} t^4 dt$$ $$\frac{(\sqrt{1+a^2})^5}{5}$$

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$\int_0^1 s \sqrt{q+s^2} \, dq=\left[\frac{2}{3} s \left(q+s^2\right)^{3/2}\right]_0^1=\frac{2}{3} s \left(\left(s^2+1\right)^{3/2}-s^3\right)$

An then

$\int_0^a \dfrac{2}{3} s \left(\left(s^2+1\right)^{3/2}-s^3\right) \, ds=\left[\dfrac{2}{15} \left(\left(s^2+1\right)^{5/2}-s^5\right)\right]_0^a=\dfrac{2}{15} \left[\left(a^2+1\right)^{5/2}-a^5-1\right]$

Hope this is useful

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