# $\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$. Is there another way to solve it faster?

I have to calculate this integral:

\begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx \qquad\text{with} \quad a \in \mathbb{R} \end{align}

My attempt:

Using, trigonometric substitution

\begin{align} \tan \theta &= \frac{x}{a}\\ \Longrightarrow \ x&=a \tan \theta\\ \Longrightarrow \ dx&=a \sec^{2}\theta\\ \Longrightarrow \ x^{2}&=a^{2}\tan^{2}\theta \end{align}

Thus, \begin{align} \int x^{2}\sqrt{a^{2}+x^{2}}\,dx&=\int a^2 \tan^{2}\theta \sqrt{a^2+a^2\tan^{2}\theta}\ a\sec^{2}\theta\, d \theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(1+\tan^{2}\theta)}\sec^{2}\theta\, d\theta\\&=a^{3}\int \tan^{2}\theta \sqrt{a^{2}(\sec^{2}\theta)}\sec^{2}\theta \, d\theta\\&=a^{4}\int (1-\sec^{2}\theta)\sec^{3}\theta \, d\theta\\&=a^{4}\underbrace{\int \sec^{3}\theta \, d\theta}_{\text{solve by parts}}-a^{4}\underbrace{\int \sec^{5}\theta \, d\theta}_{\text{solve by parts}} \end{align}

My doubt is: Is there any other way to solve it faster? Because by parts is a large process to solve each one. I really appreciate your help

• Can you use the power reduction formulas for secant integrals? It's still integration by parts, but it would speed things up. Jan 6, 2021 at 20:00
• This is fast, don't complain.
– user65203
Jan 6, 2021 at 20:57

$$x^2 \sqrt{a^2+x^2};\;\, x\to a \sinh u$$ $$dx=\cosh u\,du$$ $$\int x^2 \sqrt{a^2+x^2}\,dx=\int (a^2 \sinh ^2 u)(a \cosh u )\sqrt{a^2 \sinh ^2 u+a^2}\,du=$$ $$=a^4\int \sinh^2 u\cosh^2 u\,du=\frac{a^4}{4}\int\sinh^2 2u\,du=\frac{a^4}{8} \int (\cosh 4 u-1) \, du=$$ $$=\frac{1}{8} a^4 \left(\frac{1}{4} \sinh 4 u-u\right)+C=\frac{1}{8} \left(x \sqrt{a^2+x^2} \left(a^2+2 x^2\right)-a^4 \text{arcsinh}\left(\frac{x}{a}\right)\right)+C$$

Useful formulas

$$\cosh^2 u -\sinh^2 u=1$$

$$\sinh 2u=2\sinh u\cosh u$$

$$\cosh 4u =\sinh ^2 2u +\cosh ^2 2u=2\sinh^2 2u+1\to \sinh^2 2u = \frac{1}{2}(\cosh4u - 1)$$

$$x=a\sinh u\to u=\text{arcsinh}\left(\frac{x}{a}\right)$$

$$\sinh 4u = 2\sinh 2u \cosh 2u = 4\sinh u\cosh u(\cosh^2 u+ \sinh^2 u)$$

Integrate by parts

\begin{align} I & =\int x^{2}\sqrt{a^{2}+x^{2}}\,dx =\int \frac x3d(a^2+x^2)^{3/2} \\ &=\frac13x( a^2+x^2)^{3/2}-\frac13I-\frac{a^2}3\underset{=J}{\int \sqrt{a^2+x^2}dx} \\ &=\frac14x( a^2+x^2)^{3/2}-\frac{a^2}4J\tag 1 \end{align} Integrate $$J$$ by parts again \begin{align} J & =\int \sqrt{a^{2}+x^{2}}\,dx=x\sqrt{ a^2+x^2}-J-\int \frac{a^2}{\sqrt{a^2+x^2}}dx\\ &=\frac12 x\sqrt{ a^2+x^2}-\frac{a^2}2\sinh^{-1}\frac xa \tag2 \end{align} Plug (2) into (1) to obtain $$I= \frac14x( a^2+x^2)^{3/2}- \frac{a^2}8 x\sqrt{a^2+x^2} - \frac{a^4}8\sinh^{-1}\frac xa +C$$

HINT.-You do have a binomial integral $$\int x^{2}\sqrt{a^{2}+x^{2}}\,dx$$ in which the second case of Chebyshev is fulfilled, i.e. $$\dfrac{2+1}{2}+\dfrac12$$ is an integer for which we must have the change of variable $$u=\left(\frac{a^2+x^2}{x^2}\right)^{\dfrac12}$$ so we have after calculations the integral $$-\frac{a^5}{2}\int\frac{d(u^2-1)}{(u^2-1)^4}=\frac{a^5}{2}(u^2-1)^{-3}+C$$ NOTE.-Being in a hurry the calculation has been wrong maybe but the methode is correct (I did $$x^2=\dfrac{a^2}{u^2-1}$$ so the integral becomes $$\int\dfrac{a^2}{u^2-1}\sqrt{\frac{u^2a^2}{u^2-1}}\left(\frac{-a^2 }{(u^2-1)^2}\right)\frac{1}{\sqrt{u^2-1}}\,du$$