waigaK

2021-08-12

Evaluate the following problem.
$\int {x}^{3}{\left({a}^{2}+{x}^{2}\right)}^{\frac{1}{2}}dx$

Jayden-James Duffy

Our Aim is to evaluate the integral given below:
$\int {x}^{3}\left(\sqrt{{a}^{2}+{x}^{2}}\right)dx-\left(i\right)$
Considering the integral given below:
$\int {x}^{3}\left(\sqrt{{a}^{2}+{x}^{2}}\right)dx-\left(i\right)$
For the integrand ${x}^{2}\sqrt{{a}^{2}+{x}^{2}}$, substitute $u={x}^{2}$ and $du=2xdx$
$=\frac{1}{2}u\sqrt{{a}^{2}+u}du$
For, the integrand $u\sqrt{{a}^{2}+u}$, substitute $s={a}^{2}+u$ and ds=du
$\frac{1}{2}\int \sqrt{s}\left(s-{a}^{2}\right)ds$
Expanding the integral $\sqrt{s}\left(s-{a}^{2}\right)$ given ${s}^{\frac{3}{2}}-{a}^{2}\sqrt{s}$
$=\frac{1}{2}\int \left({s}^{\frac{3}{2}}-{a}^{2}\sqrt{s}\right)ds$
Integrate the sum term by term and taking out constants:
$=\frac{1}{2}\int {s}^{\frac{3}{2}}ds-\frac{{a}^{2}}{2}\int \sqrt{s}ds$
The integral of ${s}^{\frac{3}{2}}$ is $\frac{2{s}^{\frac{5}{2}}}{5}$
$=\frac{{s}^{\frac{5}{2}}}{5}-\frac{{a}^{2}}{2}\int \sqrt{s}ds$
The integral of $\sqrt{s}$ is $\frac{2{s}^{\frac{3}{2}}}{3}$
$=\frac{{s}^{\frac{5}{2}}}{5}-\frac{1}{3}{a}^{2}{s}^{\frac{3}{2}}+C$
Substitute back for $s={a}^{2}+u$
$=\frac{1}{5}{\left({a}^{2}+{u}^{2}\right)}^{\frac{5}{2}}-\frac{1}{3}{a}^{2}{\left({a}^{2}+u\right)}^{\frac{3}{2}}+C$
Substitute back for $u={x}^{2}$

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