I was writing an exam for my students, and I wanted to give them a trig sub integral that wasn’t too difficult (there’s this infinitesimal sweet spot between ridiculously too difficult and utterly trivial). I noticed something about virtually every single trig sub integral that I imagined or ran into – almost all of them could be solved without any trig.

Case in point: consider $$\int \frac{x^3}{\sqrt{x^2 + 1}} \, dx$$

Standard trig sub methods have us use $$x = \tan(t)$$ since $$x^2 + 1$$ appears in our integral. Making said substitution, we solve the integral as follows:

\begin{align*} \int \frac{x^3}{\sqrt{x^2 + 1}} \, dx &= \int \frac{\tan^3(t)}{\sqrt{\tan^2(t) + 1}} \, dx & x &= \tan(t) \\ &= \int \frac{\tan^3(t)}{\sqrt{\tan^2(t) + 1}} \sec^2(t) \, dt & dx &= \sec^2(t) \, dt \\ &= \int \frac{\tan^3(t)\sec^2(t)}{\sec(t)} \, dt \\ &= \int \tan^3(t)\sec(t) \, dt \\ &= \int (\sec^2(t) - 1)\sec(t)\tan(t) \, dt \\ &= \int (u^2 - 1)\sec(t)\tan(t) \, dt & u &= \sec(t) \\ &= \int u^2 - 1 \, du & du &= \sec(t)\tan(t) \, dt \\ &= \frac{1}{3} u^3 - u + C \\ &= \frac{1}{3} \sec^3(t) - \sec(t) + C \end{align*}

To simplify further, we need to draw a right triangle that satisfies $$x = \tan(t)$$, and from that triangle deduce that $$\sec(t) = (x^2 + 1)^{1/2}$$, giving us

$\int \frac{x^3}{\sqrt{x^2 + 1}} \, dx = \frac{1}{3} (x^2 + 1)^{3/2} - (x^2 + 1)^{1/2} + C$

Sometimes, you just gotta do what you gotta do, and so this is all well and good in the name of science. But what if I told you there was a way to solve this integral that involved no trig whatsoever?

The approach we can use is back substitution. Instead of setting $$x = \tan(t)$$, we’ll let $$u = x^2 + 1$$.

\begin{align*} \int \frac{x^3}{\sqrt{x^2 + 1}} \, dx &= \int \frac{x^3}{\sqrt{u}} \, dx & u &= x^2 + 1 \\ &= \frac{1}{2} \int \frac{x^2}{\sqrt{u}} \, du & du &= 2x \, dx \\ &= \frac{1}{2} \int \frac{u - 1}{\sqrt{u}} \, du & x^2 &= u - 1 \\ &= \frac{1}{2} \int \frac{u}{\sqrt{u}} - \frac{1}{\sqrt{u}} \, du \\ &= \frac{1}{2} \int u^{1/2} - u^{-1/2} \, du \\ &= \frac{1}{2} \left( \frac{2}{3}u^{3/2} - 2u^{1/2}\right) + C \\ &= \frac{1}{3}u^{3/2} - u^{1/2} + C \\ &= \frac{1}{3}(x^2 + 1)^{3/2} - (x^2 + 1)^{1/2} + C \end{align*}

At first I thought that this must simply be a fluke coincidence, but as I looked at more problems, this method kept working.

It seems as thought we can get every integral that is traditionally taught using trig substitution by back substitution instead, so long as we just remember a few basic templates:

\begin{align*} \int \frac{1}{\sqrt{A^2 - (Bx)^2}} \, dx &= \frac{1}{AB}\sin^{-1}(Bx/A) + C \\ \int \frac{1}{A^2 + (Bx)^2} \, dx &= \frac{1}{AB}\tan^{-1}(Bx/A) + C \\ \int \frac{1}{Bx\sqrt{(Bx)^2 - A^2}} \, dx &= \frac{1}{AB}\sec^{-1}(Bx/A) + C \end{align*}

Frankly, we don’t even have to remember the templates, since we can get them with the $$u$$-substitution $$u = Bx$$ and a little bit of algebra.

Now, trig is hard for students to grasp. It takes a lot of time away from what could otherwise be used to teach quantitative reasoning, logic, statistics, etc. On top of that, calculators and computer algebra systems do all of your trig for you. This simply motivates the question:

Why do we keep putting so much effort into teaching trig when a calculator can do it better than a person and the time could be used on something more relevant?

For a long time, I thought the answer was obvious. We need to teach trigonometry because once you get into Calc II, you’ll have some integrals that you’ll need trig to solve, and by then it’d be too late to start teaching trig. But seeing how trig-substitution is basically unnecessary as an integration technique, I have to wonder – why do we even teach trig-sub? And if we don’t need trigonometry for trig-sub, then why do we even still teach trigonometry at all?

We could realign the entire Mathematics curriculum. Teach just the basics of how to evaluate trig functions using a calculator or a table, and make advanced trig a special topics course for the few people who do actually still need to understand it really well. We could free up hundreds of hours for more relevant material in the process.

Now, I’m not really quite that radical, so I’m interested to hear in the comments why trigonometry (beyond a basic understanding of how to evaluate trig functions and what they tell us) should stay part of the core Calculus curriculum (maybe those of you who teach/do Physics know why). Or if you can think of an integral that actually does require trig-sub to solve (aside from the trivial integration technique of guessing the solution and checking by differentiating) I’d be interested in seeing it.