Evaluating $\int x^2\sqrt{x^2+a^2}dx$ without Trig Substitution

Question

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

What are the different ways besides trigonometric substitution that can be used to solve this problem? Moreover what would you rate this problem on a scale of toughness from 1(easy) to 10(tough)?

Answer 1

There is a sneaky substitution,$$ x=\frac{y^2-a^2}{2y} \quad\text{so}\quad y=x+\sqrt{x^2+a^2} \quad\text{and}\quad dy = { {\frac {x+\sqrt {{a}^{2}+{x}^{2}}}{\sqrt {{a}^{2}+{ x}^{2}}}}} dx. $$ Then $$ \int x^2\sqrt{x^2+a^2}dx = \int\left(\frac{a^8}{16y^5}-\frac{a^4}{8y}+\frac{y^3}{16}\right)dy =-\frac{a^8}{64 y^4}-\frac{a^4\log(y)}{8}+\frac{y^4}{64} \\={\frac {x}{4} \left( {a}^{2}+{x}^{2} \right) ^{{3/2}}}-{\frac {{a}^{2}x}{8}\sqrt {{a}^{2}+{x}^{2}}}-{\frac {{a}^{4}}{8}\log \left( x +\sqrt {{a}^{2}+{x}^{2}} \right) } $$

Answer 2

Not sure if a hyper trig substitution counts as not being a trig substitution, but it leads to a very straightforward evaluation using hyper trig half and double angle identities: \begin{eqnarray} \int x^2\sqrt{x^2+a^2}dx &=& a^4\int\sinh^2\eta\sqrt{1+\sinh^2\eta}\cosh\eta \,d\eta = a^4\int \sinh^2\eta\cosh^2\eta\,d\eta \\&=& \frac{a^4}{4}\int\sinh^2(2\eta)d\eta = \frac{a^4}{8}\int [\cosh(4\eta) -1] d\eta \\ &=& \frac{a^4}{8}\left[\frac{\sinh(4\eta)}{4} -\eta\right] = \frac{a^4}{32}\sinh\left(4\sinh^{-1}\frac{x}{a}\right) - \frac{a^4}{8}\sinh^{-1}\frac{x}{a} \end{eqnarray} Now honestly you could just leave it here, since this is a perfectly good closed form. However, if you want to get rid of the hyper trig functions, we can use the double angle identities again to get \begin{eqnarray} \sinh(4\eta) &=& 2\sinh(2\eta)\cosh(2\eta) = 4\sinh\eta\cosh\eta(1+2\sinh^2\eta) \\&=& 4\frac{x}{a}\left(1+2\frac{x^2}{a^2}\right)\sqrt{1+\frac{x^2}{a^2}} = \frac{4}{a^4}x(a^2+2x^2)\sqrt{a^2+x^2}. \end{eqnarray} Combine with $\sinh^{-1}z = \ln(z +\sqrt{1+z^2})$ to get $$ \int x^2\sqrt{x^2+a^2}= \frac{x}{8}(a^2+2x^2)\sqrt{a^2+x^2} - \frac{a^4}{8}\ln\left[\frac{x+\sqrt{x^2+a^2}}{a}\right], $$ which you can verify is equivalent to the other answers given.

In general with these sorts of quadratic integrals, there are both trig and hyper trig options for getting rid of the radicals. If unsure, you should check both, as it's often the case that one of the two will get an integrand with fewer or no trig or hyper trig functions in the denominator. This usually makes for an easier integral.

Answer 3

For terms like $\sqrt {x^2+a^2}$, the most efficient substitution is $x=a\tan\theta$.

Let $\mathcal I=\int x^2 \sqrt {x^2+a^2}\mathrm dx$.

Put $x=a\tan\theta$ and $\mathrm dx=a\sec^2\theta\mathrm d\theta$ and use $1+\tan^2\theta = \sec^2\theta$:

$$\mathcal I=\int a^2\tan^2\theta \sqrt {a^2(1+\tan^2\theta)}\,a\sec^2\theta\mathrm d\theta$$ $$=a^4 \int \tan^2\theta \sec^3\theta\mathrm d\theta$$

Reuse $1+\tan^2\theta = \sec^2\theta$ and continue solving from here.

For reference, here's the final result:

$$\mathcal I=-\frac{a^4\ln\left(a\sqrt {x^2+a^2}+|a|x\right)}{8}+\frac{a\sqrt {x^2+a^2}(2x^3+a^2x)}{8|a|}+C$$

Answer 4

Let $x=a\tan\theta$. Then, by integration by parts, $\begin{align} I&= \int x^2\sqrt{x^2+a^2}dx\\ &=a^4\int\tan^2\theta\sec^3\theta\,d\theta\\ &=\frac{a^4}{3}\tan^3\theta\sec\theta-\frac{a^4}{3}\int\tan^4\theta\sec\theta\,d\theta\\ &=\frac{a^4}{3}\tan^3\theta\sec\theta +\frac{a^4}{3}\int\tan^2\theta\sec\theta\,d\theta-\frac13 I\\ \end{align}$ Hence, $$I=\frac{a^4}{4}(\tan^3\theta\sec\theta+\int\sec^3\theta\,d\theta-\int\sec\theta\, d\theta)$$ Here, we will use this well-known result. Then, $$I=\frac{a^4}{4}(\tan^3\theta\sec\theta+\frac12\tan\theta\sec\theta-\frac12\ln|\tan\theta+\sec\theta|)+c$$ and $$I=\frac{1}{4}x^3\sqrt{x^2+a^2}+\frac18 a^2x\sqrt{x^2+a^2}-\frac18 a^4\ln|x+\sqrt{x^2+a^2}|)+c$$

Answer 5

Taking $x$ inside the square root,

$$\int x^2\sqrt{x^2+a^2}\mathrm dx$$ $$=\frac{\text{sgn}x}2\int\sqrt{x^4+a^2x^2}\mathrm d(x^2)$$

Substituting $t=x^2$, the integral becomes

$$\frac{\text{sgn}x}2\int\sqrt{\left(t+\frac{a^2}2\right)^2-\left(\frac{a^2}2\right)^2}\mathrm dt$$

Now, you can finish off by using the standard integral formula

$$\int\sqrt{x^2-a^2}\mathrm dx=\frac{x}2\sqrt{x^2-a^2}-\frac{a^2}2\ln\left|x+\sqrt{x^2-a^2}\right|+C$$