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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 $$ K_3(\theta) = \dfrac{1}{2}\sec \theta \tan \theta + \dfrac{1}{2}\ln(\sec \theta + \tan \theta) $$
Hint: Use differential binomial and the following substitution:
$$1+x^2=t^2x^2$$
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:
$$\int x^2\sqrt{x^2-1}\,dx=\frac{1}{4}x^4\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}x^2\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}\ln\bigg|x+\sqrt{1-\frac{1}{x^2}}\bigg|+C$$
