2
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ What have you tried so far? What integration techniques seem like they might be helpful? $\endgroup$ – Semiclassical Jul 30 '14 at 1:15
  • $\begingroup$ $u=x^2-1$ is a solid substitution here. $\endgroup$ – Adam Hughes Jul 30 '14 at 2:05
4
$\begingroup$

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 $$ K_3(\theta) = \dfrac{1}{2}\sec \theta \tan \theta + \dfrac{1}{2}\ln(\sec \theta + \tan \theta) $$

$\endgroup$
  • $\begingroup$ In the second $\int$, you are missing a factor of $\sec\theta\tan\theta$ before ${\rm d}\theta$. $\endgroup$ – Felix Marin Jul 30 '14 at 1:34
2
$\begingroup$

Hint: Use differential binomial and the following substitution:

$$1+x^2=t^2x^2$$

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$$ \begin{aligned}\int x^2\sqrt{x^2 - 1}\,\,\mathrm{d}x&=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{x^4-1+1}{\sqrt{x^2 - 1}}\,\mathrm{d}x\\ &=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{\mathrm{d}x}{\sqrt{x^2 - 1}} - \frac{1}{3}\int\frac{(x^2-1)(x^2+1)}{\sqrt{x^2-1}}\,\mathrm{d}x\\ &=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{\mathrm{d}x}{\sqrt{x^2 - 1}} - \frac{1}{3}\int(x^2+1)\sqrt{x^2 - 1}\,\mathrm{d}x\\ &=\tfrac{1}{3}x^3\sqrt{x^2 - 1} - \frac{1}{3}\int\frac{\mathrm{d}x}{\sqrt{x^2 - 1}} -\frac{1}{3}\int x^2\sqrt{x^2-1}\,\mathrm{d}x - \frac{1}{3}\int \sqrt{x^2-1}\,\mathrm{d}x\\ &=\tfrac{1}{4}x^3\sqrt{x^2-1}-\frac{1}{4}\int\frac{\mathrm{d}x}{\sqrt{x^2-1}}-\frac{1}{4}\int\sqrt{x^2 - 1}\,\mathrm{d}x\\ &=\tfrac{1}{4}x^3\sqrt{x^2-1}-\frac{1}{4}\ln\left|x +\sqrt{x^2-1}\right| - \frac{1}{4}\int\sqrt{x^2-1}\,\mathrm{d}x \end{aligned} $$

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, $$ \int x^2 \sqrt{x^2 - 1}\,\mathrm{d}x = \frac{x}{4}\!\!\left(x^2-\frac{1}{2}\right)\!\sqrt{x^2 - 1} - \frac{1}{8}\ln\left|x+\sqrt{x^2-1}\right| + C $$

$\endgroup$
0
$\begingroup$

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:

$$\int x^2\sqrt{x^2-1}\,dx=\frac{1}{4}x^4\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}x^2\sqrt{1-\frac{1}{x^2}}-\frac{1}{8}\ln\bigg|x+\sqrt{1-\frac{1}{x^2}}\bigg|+C$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.