How do I evaluate the following indefinite integral?
$$\int x^2 \sqrt{x^2-1} dx$$
Through integration of parts, I have obtained
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int (x^2-1)^{3/2} dx $$
I've attempted evaluating the second term through substitution, where
$$ x = \sec(u)$$
However, I am stuck with
$$ \frac{x}{3}(x^2-1)^{3/2} - \frac{1}{3} \int \tan^4(u) \sec (u) du $$
What would be my next step?
Let $x = \sec \theta$. Hence, $dx = \sec \theta \tan \theta d\theta$. $$ \int x^2\sqrt{x^2 - 1}dx = \int \sec^2\theta \tan \theta \sec \theta \tan \theta d\theta = \int \sec^3 \theta \tan^2 \theta d\theta $$ $$ = \int \sec^3d\theta(\sec^2\theta - 1)d\theta = \int \sec^5\theta d\theta - \int \sec^3 \theta d\theta $$ In this link $$ K_n(\theta) := \int \sec^n \theta d\theta = \frac{1}{n-1}\sec^{n-2}\theta\tan \theta + \frac{n-2}{n-1}K_{n-2}(\theta) $$ Thus, $$ \int \sec^5\theta d\theta = \dfrac{1}{4}\sec^3d\theta \tan \theta d\theta + \dfrac{3}{4}K_3(\theta) $$ and $$ K_3(\theta) = \dfrac{1}{2}\sec \theta \tan \theta + \dfrac{1}{2}\ln(\sec \theta + \tan \theta) $$
How do I evaluate the following indefinite integral?
As a general rule, whenever evaluating an integral containing $\sqrt{x^2\pm a^2}$, one of the most natural substitutions is $x=a\cosh t$ or $x=a\sinh t$, depending on the sign. Useful formulas are $\cosh^2t-\sinh^2t=1,~\cosh't=\sinh t,~\sinh't=\cosh t,~\sinh(2t)=2\sinh t\cosh t,~$ etc.
$$ \begin{aligned}\int x^2\sqrt{x^2 - 1}\,\,\mathrm{d}x&=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{x^4-1+1}{\sqrt{x^2 - 1}}\,\mathrm{d}x\\ &=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{\mathrm{d}x}{\sqrt{x^2 - 1}} - \frac{1}{3}\int\frac{(x^2-1)(x^2+1)}{\sqrt{x^2-1}}\,\mathrm{d}x\\ &=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{\mathrm{d}x}{\sqrt{x^2 - 1}} - \frac{1}{3}\int(x^2+1)\sqrt{x^2 - 1}\,\mathrm{d}x\\ &=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{\mathrm{d}x}{\sqrt{x^2 - 1}} -\frac{1}{3}\int x^2\sqrt{x^2-1}\,\mathrm{d}x - \frac{1}{3}\int \sqrt{x^2-1}\,\mathrm{d}x\\ &=\tfrac{1}{4}x^3\sqrt{x^2-1}-\frac{1}{4}\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}-\frac{1}{4}\int\sqrt{x^2 - 1}\,\mathrm{d}x\\ &=\tfrac{1}{4}x^3\sqrt{x^2-1}-\frac{1}{4}\ln\left|x +\sqrt{x^2-1}\right| - \frac{1}{4}\int\sqrt{x^2-1}\,\mathrm{d}x \end{aligned} $$
To evaluate $\displaystyle{\int\sqrt{x^2 - 1}\,\mathrm{d}x}$, use integration by parts again: $$ \begin{aligned} \int\sqrt{x^2 - 1}\,\mathrm{d}x&=x\sqrt{x^2 - 1} -\int\frac{x^2-1+1}{\sqrt{x^2-1}}\,\mathrm{d}x\\ &=x\sqrt{x^2 - 1} - \int\sqrt{x^2-1}\,\mathrm{d}x-\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}\\ &=\tfrac{1}{2}x\sqrt{x^2-1}-\tfrac{1}{2}\ln\left|x+\sqrt{x^2 - 1}\right|+C_{0} \end{aligned} $$ Thus, $$ \int x^2 \sqrt{x^2 - 1}\,\mathrm{d}x = \frac{x}{4}\!\!\left(x^2-\frac{1}{2}\right)\!\sqrt{x^2 - 1} - \frac{1}{8}\ln\left|x+\sqrt{x^2-1}\right| + C $$
Used a different version of the secant reduction formula. I really enjoyed this problem!
Let $x=\sec(u)$ and $dx=\sec(u)\tan(u)\,dx$ so that:
$$\int x^2\sqrt{x^2-1}\,dx=\int\sec^2(u)\sqrt{\sec^2(u)-1}\sec(u)\tan(u)\,du$$
Now simplify, noting that $\sec^2(u)-1=\tan^2(u)$:
$$=\int\sec^3(u)\tan^2(u)\,du$$
And again, using $\sec^2(u)-1=\tan^2(u)$ and simplifying:
$$\int(\sec^5(u)-\sec^3(u))\,du$$
Apply the version of the secant reduction formula containing tangent, which is:
$$\int\sec^n(u)\,du=\frac{\tan(u)\sec^{n-2}(u)}{n-1}+\frac{n-2}{n-1}\int\sec^{n-2}(u)\,du$$
To get:
$$\int(\sec^5(u)-\sec^3(u))\,du=\frac{\tan(u)\sec^3(u)}{4}+\frac{3}{4}\bigg(\frac{\tan(u)\sec(u)}{2}+\frac{1}{2}\ln|\sec(u)+\tan(u)|\bigg)$$ $$-\bigg(\frac{\tan(u)\sec(u)}{2}+\frac{1}{2}\ln|\sec(u)+\tan(u)|\bigg)$$
Next apply the substitutions $x=\sec^{-1}(u)$ and $\tan(\sec^{-1}(u)=x\sqrt{1-\frac{1}{x^2}}$ and simplify:
$$\int x^2\sqrt{x^2-1}\,dx=\frac{1}{4}x^4\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}x^2\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}\ln\bigg|x+\sqrt{1-\frac{1}{x^2}}\bigg|+C$$
For the domain $|x|>1$,
\begin{align} \int x^2 \sqrt{x^2-1}\ dx=& \int\frac{x}{4\sqrt{x^2-1}}d\left[(x^2-1)^2\right]\\ \overset{ibp}=&\ \frac14 x(x^2-1)^{3/2} +\frac18\int \frac{\sqrt{x^2-1}}x d(x^2)\\ \overset{ibp}= &\ \frac14 x(x^2-1)^{3/2}+\frac18 x(x^2-1)^{1/2}-\frac18 \tanh^{-1}\frac x{\sqrt{x^2-1}} \end{align}
$$\int x^{2}\sqrt{x^{2}-1}\,dx\\ x=\cosh{y}\Rightarrow dx=\sinh{y}dy\\ \int x^{2}\sqrt{x^{2}-1}\,dx=\int \cosh^{2}{y}\sinh^{2}{y}\,dy=\frac{1}{4}\int\sinh^{2}{2y}\,dy\\ =\frac{1}{4}\int\frac{\cosh{4y}-1}{2}\,dy=\frac{\sinh{4y}}{2}-\frac{y}{8}+C\\ =\sinh{2y}\cosh{2y}-\frac{\cosh^{-1}x}{8}+C\\ =2(\cosh{y})(\cosh{2y})(\sinh{y})-\frac{\cosh^{-1}x}{8}+C\\ =2x(2x^{2}-1)\sqrt{x^{2}-1}-\frac{\cosh^{-1}x}{8}+C\\ $$
Why not exponential substitution? We apply the following general transformation formula
$$\int f\left(x, \sqrt{x^2-a^2}, \frac{\sqrt{x-a}}{\sqrt{x+a}}\right)\,dx= \int f\left(\frac{e^{i\alpha}+e^{-i\alpha}}{2}a, \frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2}a, \frac{1-e^{\pm\text{i}\alpha}}{1+e^{\pm\text{i}\alpha}}\right)\,\frac{e^{\mp\text{i}\alpha}-e^{\pm\text{i}\alpha}}{2i}a\,d\alpha. \tag{1}$$
Where $\alpha=\cos^{-1}(\frac{x}{a})$ and for the alternating signs $\mp$, use the upper sign when $\frac{x}{a} \geq 1$, and the lower sign when $0 \leq \frac{x}{a} \leq 1$.
After applying $(1)$ the integral becomes
$$\begin{aligned}\int x^2\sqrt{x^2-1}\,dx&=-\frac{i}{16}\int \left( e^{i\alpha} + e^{-i\alpha} \right)^2 \left( e^{-i\alpha} - e^{i\alpha} \right)^2 \, d\alpha\\&=-\frac{i}{16}\int \left(e^{-4 i \alpha} + e^{4 i \alpha} - 2 \right)\,d\alpha\\&=\frac{1}{64}e^{-4i\alpha} - \frac{1}{64} e^{4i\alpha} +\frac{i\alpha}{8} + C\\&=\frac{i}{32}(4\alpha - \sin(4\alpha)) + C\\&= \frac{i}{32}(4\cos^{-1}(x) - \sin(4\cos^{-1}(x))) + C\\&=\frac{1}{8}\left(\ln|x-\sqrt{x^2-1}| +x \left(2x^2-1\right) \sqrt{x^2-1}\right) + C \end{aligned}$$
