# Why is this secant substitution allowed?

On Paul's Math Notes covering Trig Substitutions for Integrals we start with an integral:

$$\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}}$$

Right away he says to substitute $$x=\frac{2}{5}\sec(θ)$$. Why is that allowed?

Looking further down onto how he approaches the problem, it seems like it's allowed because it's compensated for with a dx:

$$dx = \frac{2}{5}\sec \theta \tan \theta \,d\theta$$

Is that what's going on here? It's fair to say you can substitute x with whatever you want so long as you update dx? Seems like it wouldn't work for constant functions of x, like $$x = 5$$.. since that'd get you $$dx=0$$ and clearly be wrong. So what rules are in play here for substitution?

• Perhaps this is relevant Jan 27, 2022 at 8:21

The exact rule is: if $$x(t)$$ is a differentiable function of $$t$$, then for any continuous integrand $$f(t)$$, $$\int f\big( x(t) \big) x'(t)\,dt = \bigg( \int f(x)\,dx \bigg) \bigg| _{x=x(t)},$$ where the subscript on the right-hand side means that after the indefinite integral $$\int f(x)\,dx$$ is evaluated, one then plugs in $$x(t)$$ for $$x$$. This can be proven easily by differentiating both sides with respect to $$t$$—the right-hand side is a composition of two functions, with the inner function being $$x(t)$$, by the definition of the notation.

You're interested in the situation where we start from the right-hand side and choose the function $$x(t)$$ (as opposed to the more standard substitution method where one starts from the left-hand side); in particular you're asking what happens if you set $$x=5$$ for example. Then the left-hand side is $$\int 0\,dt = C$$, while if $$F(x)$$ is an antiderivative of $$f(x)$$ then the right-hand side is $$\big(F(x)+C'\big)\big|_{x=5}$$, which is simply $$F(5)+C'$$. Since $$C$$ and $$C'$$ represent arbitrary constants and $$F(5)$$ is some constant, these two answers represent the same family of functions. It's not wrong after all—just not useful!

• Going to grok this.. one nit though-- it seems you have a floating comma on your bar in your equation and I'm not sure if that's intentional. I've only seen the bar notation used to indicate you substract things.. I guess this is a nother use? Jan 27, 2022 at 6:51
• @bgcode It's intentional because mathematical expressions should be read as a part of a sentence and how MathJax works. If the comma was placed after , it will be placed below, which is awkward. Jan 27, 2022 at 6:59
• This reminds me vaguely of the morning I was disturbed by the thought that the Fourier coefficients of the rational indicator function were all 0 and I was concerned because they form a Hilbert basis....I had completely forgotten that in $L^2$ space, $0$ and the rational indicator function are the same :)
– Alan
Jan 27, 2022 at 7:28
• Ok cool. Is the bar here the same as the bar used to minus things when applying FToC? Or this has completely different meaning? Jan 27, 2022 at 8:15
• It has a different but related meaning—both are "plug the stuff to my right into the expression on my left". I don't love the notation here, to be honest, but I can't think of a better one without introducing a new function $F(x)$ for an antiderivative of $f(x)$. Jan 27, 2022 at 18:16

To add some information to Greg Martin's answer, this is because in some cases, integrals with $$\sqrt{\pm(ax)^2 \pm b^2}$$ can be solved using trigonometric substitutions. Recall the identities $$\sin^2\theta + \cos^2\theta = 1, \\ 1 + \cot^2\theta = \csc^2\theta, \\ \tan^2\theta + 1 =\sec^2\theta.$$

In the case $$\sqrt{(ax)^2 - b^2}$$, we can factor out $$b^2$$ from the square root, giving us $$b\sqrt{(\frac{a}{b}x)^2 -1}$$. By comparing the term inside the square root to the trig identities, it looks the same to $$\csc^2\theta - 1$$ and $$\sec^2\theta - 1$$. Hence, we can use $$\frac{a}{b}x = \csc\theta \Leftrightarrow x = \frac{b}{a}\csc\theta$$ or $$\frac{a}{b}x = \sec\theta \Leftrightarrow x = \frac{b}{a}\sec\theta$$.

Let's try to solve $$\int{{\frac{{\sqrt {25{x^2} - 4} }}{x}\,dx}}$$ using the substitution $$x = \frac{2}{5}\csc\theta$$. Then, $$dx = -\frac{2}{5}\csc\theta\cot\theta$$. Also, $$\sqrt{25x^2 - 4}$$ becomes $$2\cot\theta$$ assuming that $$\cot\theta > 0$$. This gives us $$\int \frac{2\cot\theta}{\frac{2}{5}\csc\theta}\left(-\frac{2}{5}\csc\theta\cot\theta\right) \\ -2\int\cot^2\theta\,d\theta \\ -2\int(\csc^2\theta - 1)\,d\theta \\ -2(-\cot\theta - \theta) + C \\ 2(\cot\theta + \theta) + C \\ 2\left(\frac{\sqrt{25x^2 - 4}}{2} + \sin^{-1}\left(\frac{2}{5x}\right)\right) + C \\ \sqrt{25x^2 - 4} + 2\sin^{-1}\left(\frac{2}{5x}\right) + C.$$

Comparing the answer obtained when $$x = \frac{2}{5}\sec\theta$$: $$\sqrt {25{x^2} - 4} - 2{\cos ^{ - 1}}\left( {\frac{2}{{5x}}} \right) + C$$ to the substitution $$x = \frac{2}{5}\csc\theta$$, we can see that for all $$x \in \mathbb{R}$$, $$\sin^{-1}\left(\frac{2}{5x}\right) = -\cos^{-1}\left(\frac{2}{5x}\right) + C$$ will hold if and only if $$C = \frac{\pi}{2}$$. By substituting values of $$x$$, we can see that it is indeed the case.

You need to remember the Pythagorean trigonometric identity that says $$\sec^2 \theta-1 = \tan^2\theta.$$

Where you see $$\Big( \big(\text{variable}\big)^2 - \text{positive constant} \Big),$$ you can often use this substitution in just the way in which it is used here.