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Right off the bat I factored a $4$ from the radicand to get it into a form such that I can leverage $${\tan^2(\theta)} = {\sec^2(\theta)} - 1$$

Then I set

$$\begin{align*} \frac{25}{4} x^2 &= \sec^2(\theta) \\ x &= \frac{2}{5}\sec(\theta) \\ \mathrm{d}x &= \frac{2}{5} \sec(\theta)\tan(\theta) \, \mathrm{d} \theta \end{align*}$$

Now I went ahead and substituted into the integrand; after simplification and evaluation of the new integral we get

$$2\tan\theta - 2\theta$$

Now to sub back in terms of $x$ by using a right triangle $$2\tan(\theta) = \sqrt{ 25x^2 - 4}$$

(Note: on the right triangle, Opposite = $\sqrt{ 25x^2 - 4}$, Adjacent = $2$, hypotenuse = $5x$)

However for $2\theta$ I made the resubstitution that

$$ x = \frac{2}{5} \sec\theta $$

so to solve for $\theta$

$$\operatorname {arcsec} \frac{5}{2} x = \theta$$

However that was the wrong resubstitution for $\theta$; the correct one was

$$\theta = \arccos \left(\frac{2}{5x}\right)$$

How did $\arccos$ even get into the picture? We already had $x = \frac{2}{5}\sec(\theta)$ so intuitively I just solved for theta. I don't understand how it's $\arccos$ and not $\operatorname {arcsec}$; can someone help me make sense of the last part?

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3 Answers 3

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Such trig substitution is problematic in domain-signs and subsequent back-subs. Preferably, substitute $t= \sqrt{ 25x^2 - 4} $

$$\int \frac{\sqrt{ 25x^2 - 4}}{x}dx =\int\frac{t^2}{t^2+4}dt=t-2\tan^{-1}\frac t2+C $$

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$x = \frac{2}{5} \cosh \theta$ and then you get a linear combination of $\cosh\theta$ and $\mathrm{sech\,}\theta$ to integrate, both of which are standard.

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  • $\begingroup$ I dont follow. Please elaborate $\endgroup$ Commented Nov 8, 2018 at 10:52
  • $\begingroup$ Make the indicated substitution, and do the algebra. $\endgroup$ Commented Nov 8, 2018 at 10:57
  • $\begingroup$ @RichardMartin, that doesn't seem to work. Can you show the details in your answer? The radicand becomes $\sqrt{4 \sinh^2 \theta} = 2 \sinh \theta$ and $dx = (2/5) \sinh \theta d\theta$. So the integral becomes $\int 2 \sinh \theta ((2/5) \sinh \theta d\theta) / (2/5) \cosh \theta = 2 \int \sinh^2 \theta d\theta / \cosh \theta$, but I don't know any identity to give me a linear combination of $\cosh \theta$ and $sech \theta$. $\endgroup$ Commented Jul 2, 2023 at 20:22
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EDIT: A Late Addressing of OP's Actual Question:

(No clue why I - or anyone outside the comments - never thought to actually do this years ago, but whatever.)

OP's confusion lies in choosing to do an arcsecant substitution, $$ x = \frac 2 5 \sec \theta $$ whereas another source claims the substitution $$ \theta = \arccos \left( \frac{2}{5x} \right) $$

The difference here is elementary: namely, there is none. Observe that, in a simpler case, $$ z = \sec \theta \implies \frac 1 z = \frac{1}{\sec \theta} = \cos \theta $$ so $$ \theta = \operatorname{arcsec}(z) = \arccos \left( \frac 1 z \right) $$ Making either substitution is entirely valid. Some people just prefer to default to the "main three" trigonometric functions (sine, cosine, tangent) and not deal with the reciprocal functions, but there is little reason why you can't.


What follows from here is a touching up on my original post for posterity's sake. Therein, I actually simply solved the integral, not in a very helpful way for OP considering the then-absence of the above. However, I solved it incorrectly at the time, but I've corrected and checked this solution. I've also included a minor aside on "alternative" trigonometric substitutions, since my original answer hinted at such with no elaboration.


Solving The Integral With Arcsecant Substitution:

Generally, seeing $\sqrt{x^2 - a^2}$ or something of the like begs for the substitution of $x = a\sec(u)$, since $\tan^2(x) = \sec^2(x) - 1$. (Or perhaps one of several other substitutions$^{[1]}$; they follow a similar process, I'll leave that calculation up to you if you prefer.)

So, notice, by factoring out $\sqrt{25}=5$ from the numerator,

$$\newcommand{\arcsec}{\operatorname{arcsec}} \mathcal I := \int \frac{\sqrt{25x^2 - 4}} x \, \mathrm{d}x = 5 \int \frac{\sqrt{x^2 - (2/5)^2}}{x} \, \mathrm{d}x$$

Let $x = \frac 2 5 \sec(u)$. Then $\mathrm{d}x = \frac 2 5 \sec(u)\tan(u) \, \mathrm{d}u$. Notice the radicand becomes

$$x^2 - \left( \frac 2 5 \right)^2 = \left( \frac 2 5 \right)^2 \Big(\sec(u)^2 - 1 \Big) = \left( \frac{2 \tan(u)}{5} \right)^2$$

Thus,

$$\mathcal I = 5 \int \underbrace{\frac{5}{2 \sec(u)}}_{ = 1/x} \cdot \underbrace{\frac{2}{ 5} \tan(u)}_{=\sqrt{x^2 - (2/5)^2}} \cdot \underbrace{ \frac 2 5 \sec(u)\tan(u) \, \mathrm{d}u}_{=\, \mathrm{d}x}$$

Cancelling terms and grouping them yields

$$\mathcal{I} = 2 \int \tan^2(u)\, \mathrm{d}u$$

This integral is fairly standard. Note that $$ \tan^2(u) = \sec^2(u) - 1 $$ and $$ \int \sec^2(x) \, \mathrm{d}x = \tan(x) + C \qquad \int \mathrm{d}x = x + C $$ and do back-substitution to get $$ \mathcal{I} = 2 \tan \left( \arcsec \left( \frac 5 2 x \right) \right) - 2 \arcsec \left( \frac 5 2 x \right) + C $$ though we can do better and simplify the first term further.

In more detail, note that $$ x = \frac 25 \sec(u) \implies \frac{5x}{2} = \sec(u) = \frac{\text{"hypotenuse"}}{\text{"adjacent"}} $$ so, filling out a triangle with acute angle $u$, hypotenuse $5x$, and adjacent side $2$, the remaining side by the Pythagorean theorem is found to be $\sqrt{25x^2 - 4}$:

enter image description here

From here, we easily see that $$ \tan(u) = \tan \left( \arcsec \left( \frac 5 2 x \right) \right) = \frac{\sqrt{25x^2 - 4}}{2} $$

Thus, $$ \mathcal{I} = \sqrt{25x^2 - 4} - 2 \arcsec \left( \frac 5 2 x \right) + C $$


Alternative Trigonometric Substitutions$^{[1]}$:

Given integrals with certain items in them, a variety of substitutions may come into play. It is typical to teach three common ones, with the idea of leveraging the Pythagorean identities.

Those identities are, for ordinary trigonometric functions, $$\begin{align*} \sin^2 \theta +\cos^2 \theta &= 1 \tag{1} \\ \tan^2 \theta + 1 &= \sec^2 \theta \tag{2} \\ 1 + \cot^2 \theta &= \csc^2 \theta \tag{3} \end{align*}$$

The "standard" substitutions are

  • $\sqrt{1-x^2} \implies x = \sin \theta$ (leverage identity $(1)$)
  • $\sqrt{x^2 - 1} \implies x = \sec \theta$ (leverage identity $(2)$)
  • $\sqrt{1+x^2} \implies x = \tan \theta$ (leverage identity $(2)$)

Some others can also be used, if you're careful about the bounds of integration and their order:

  • $\sqrt{1-x^2} \implies x = \cos \theta$ (leverage identity $(1)$)
  • $\sqrt{x^2 - 1} \implies x = \csc \theta$ (leverage identity $(3)$)
  • $\sqrt{1+x^2} \implies x = \cot \theta$ (leverage identity $(3)$)

One likewise has identities for the hyperbolic trigonometric functions,

$$\newcommand{\sech}{\operatorname{sech}} \newcommand{\csch}{\operatorname{csch}} \begin{align*} \cosh^2 \theta - \sinh^2 \theta &= 1 \tag{4} \\ 1 - \tanh^2 \theta &= \sech^2 \theta \tag{5} \\ \coth^2 \theta - 1 &= \csch^2 \theta \tag{6} \end{align*}$$

and corresponding potential "standard" substitutions (even if not often taught)

  • $\sqrt{1+x^2} \implies x = \sinh \theta$ (leverage identity $(4)$)
  • $\sqrt{x^2 - 1} \implies x = \cosh \theta$ (leverage identity $(4)$)

and the less standard substitutions

  • $\sqrt{1-x^2} \implies x = \tanh \theta$ (leverage identity $(5)$)
  • $\sqrt{1-x^2} \implies x = \sech \theta$ (leverage identity $(5)$)
  • $\sqrt{x^2 - 1} \implies x = \coth \theta$ (leverage identity $(6)$)
  • $\sqrt{x^2 + 1} \implies x = \csch \theta$ (leverage identity $(6)$)

(Note, too, in all of these discussions, technically these square roots should naively simplify to $|\cos \theta|, |\sin \theta|,$ etc. This generally can be ignored, for reasons discussed here.)

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  • $\begingroup$ You wrote $\cos^2 u - 1 = \sin^2 u$, but the correct identity is $\cos^2 u + \sin^2 u = 1$, so $\sin^2 u = 1 - \cos^2 u$ and not $\sin^2 = \cos^2 u - 1$. That's derailing your calculation in "notice the radicand becomes...". $\endgroup$ Commented Jul 2, 2023 at 19:52
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    $\begingroup$ You're right. I've gone ahead and fixed that. (And actually edited my poor answer to actually reflect on OP's question as well, and elaborated a little on some other things.) $\endgroup$ Commented Jul 2, 2023 at 22:44
  • $\begingroup$ You did not explain why $\sqrt{(2/5) \tan u)^2} = (2/5) \tan u$. Without some justification, all we can say is that $\sqrt{(2/5) \tan u)^2} = |(2/5) \tan u|$. Can you justify that step? $\endgroup$ Commented Jul 4, 2023 at 18:40
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    $\begingroup$ This is handled in the post I link to at the bottom of the post, and mentioned explicitly in the comments of its top answer. $\endgroup$ Commented Jul 4, 2023 at 19:58

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