Calculate $\int x^2\sqrt{a^2+x^2}dx$

Question

Calculate $$\int x^2\sqrt{a^2+x^2}dx$$

My try (directly integrating by parts): $$\int x^2\sqrt{a^2+x^2}=\dfrac{x^3}{3}\sqrt{a^2+x^2}-\int \dfrac{x^3}{3}\dfrac{x}{\sqrt{x^2+a^2}} dx=\\=\dfrac{x^3}{3}\sqrt{a^2+x^2}-\dfrac13\int \dfrac{x^3\cdot x}{\sqrt{x^2+a^2}}dx=\\=\dfrac{x^3}{3}\sqrt{a^2+x^2}-\dfrac13\int x^3d\sqrt{x^2+a^2}.$$ How do we calculate this (new) integral though?

P.S. I am not yet familiar with the approach that uses substitution for solving integrals, so I'd rather integrate by parts on this occasion (as I think it would probably work).

Answer 1

You can solve it via hyperbolic substitution plus integration by parts (it'd be too cumbersome to work only with IBP).

We begin by doing $x=a\sinh\theta$, thus $dx=a\cosh\theta d\theta$. $$ \begin{aligned} \mathcal J\equiv\int x^2\sqrt{a^2+x^2}dx&=\int a^2\sinh^2\theta\sqrt{a^2+a^2\sinh^2\theta}\ a\cosh\theta d\theta\\ &=a^4\int\sinh^2\theta\cosh^2\theta d\theta\\ &=\dfrac{a^4}{4}\int\sinh^22\theta d\theta. \end{aligned}$$

Now we proceed with IBP in order to compute $\int\sinh^22\theta d\theta$.

We choose $u=\sinh2\theta$ and $dv=\sinh2\theta d\theta$, meaning that $du=2\cosh\theta d\theta$ and that $v=\frac{1}{2}\cosh2\theta \ (+C)$. $$ \begin{aligned} \frac{4}{a^4}\mathcal J=\mathcal I=\int\sinh^22\theta d\theta&=\frac{1}{2}\sinh 2\theta\cosh2\theta-\int\cosh^22\theta d\theta\\ &=\frac{1}{2}\sinh 2\theta\cosh2\theta-\int(1+\sinh^22\theta)d\theta\\ &=\frac{1}{2}\sinh 2\theta\cosh2\theta-\theta-\mathcal I \end{aligned}$$ $$\implies \mathcal I=\frac{1}{2}\sinh 2\theta\cosh2\theta-\theta-\mathcal I$$ $$\hspace{0.5cm}\implies\mathcal I=\frac{1}{4}\sinh 2\theta\cosh2\theta-\frac{\theta}{2}.$$

At last we revert the substitution expressing $\theta$ in function of $x$ $\left(\theta=\sinh^{-1}\left(\frac{x}{a}\right)\right)$. Thus, the final result should be $$\mathcal J=\frac{a^4}{16}\sinh\left[2\sinh^{-1}\left(\frac{x}{a}\right)\right]\cosh\left[2\sinh^{-1}\left(\frac{x}{a}\right)\right]-\frac{a^4}{8}\sinh^{-1}\left(\frac{x}{a}\right)+C.$$

(Maybe it can be further simplified in the form of radicals?)

Answer 2

Among substitutions, hyperbolic one may be simpler.

WLOG, let’s put $x=a\sinh \theta$ where $a>0$, then the integral is transformed into $$ \begin{aligned} I & =\int a^2 \sinh ^2 \theta \cdot a \cosh \theta \cdot a \cosh \theta d \theta \\ & =a^4 \int(\sinh \theta \cosh \theta)^2 d \theta \\ & =\frac{a^4}{4} \int \sinh ^2 2 \theta d \theta \\ & =\frac{a^4}{4} \int \frac{\cosh 4 \theta-1}{2} d \theta \\ & =\frac{a^4}{8}\left(\frac{\sinh 4 \theta}{4}-\theta\right)+C \\ & =\frac{a^4}{8}(\sinh \theta \cosh \theta \cosh 2 \theta-\theta)+C \\ & =\frac{a^4}{8}\left[\frac{x}{a} \cdot \frac{\sqrt{a^2+x^2}}{a} \cdot\left(\frac{a^2+2 x^2}{a^2}\right)-\sinh ^{-1}\left(\frac{x}{a}\right)\right]+C\\& =\frac{1}{8}\left[x \sqrt{a^2+x^2}\left(a^2+2 x^2\right)-a^4 \sinh^{-1} \left(\frac xa\right)\right]+C \end{aligned} $$ Wish it helps.

Answer 3

Integrate by parts twice as follows

\begin{align} \int x^2\sqrt{a^2+x^2}\ dx =& \int \frac{x}{4 \sqrt{a^2+x^2}}d[(a^2+x^2)^2]\\ \overset{ibp}=&\ \frac x4(a^2+x^2)^{\frac 32}-\frac{a^2}4 \int \sqrt{a^2+x^2}\ dx\\ \int \sqrt{a^2+x^2}\ dx =& \int \frac{\sqrt{a^2+x^2}}{2x}d(x^2)\\ \overset{ibp}=&\ \frac x2 \sqrt{a^2+x^2}+\frac{a^2}2 \int \frac1{\sqrt{a^2+x^2}}\ dx\\ =&\ \frac x2 \sqrt{a^2+x^2}+\frac{a^2}2\sinh^{-1}\frac xa \end{align}