A nice way to remember trigonometric integrals?

Question

Is there a "nice" way to remember trigonometric integrals, beyond what is typically taught in a standard calculus class? I'm currently in Calculus II, and up to now I've found calculus rather accessible. I love that, at least in my classes, we learn the "how" and the "why". I'm struggling, however, to remember "trigonometric integrals" on exams, etc, where notes aren't allowed. We're effectively given an integration table, and tasked with memorizing maybe 15 or 20 results in a couple week's time (no notes, and no calculator are allowed on exams, in-class quizzes, and technically calculators are not allowed on homework).

So, is there a "nice" way to remember these, beyond something like a mnemonic, etc? Perhaps some line of reasoning or a simple proof, etc? I tend to more easily remember things when I understand their derivation/intuition, if nothing else because I'm able to recreate it on the spot without memorizong the details.

Also, to clarify, by "trigonometric integrals", I'm referring to the integrals of the trigonometric functions ($\sin$, $\cos$, $\tan$, and $\sec$), the inverse trigonometric integrals ($\sin^{-1}$, etc), and integrals like:

$\int \frac{1}{x^2+1}$

...which work out to be trigonometric functions, products of trigonometric functions, etc.

Answer 1

For me, I am remembering just a few formula, and even then most of those are from derivatives. So $(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$. This allows me to put an integral sign before those and get the formula for integrals. For tangent I use integration by parts. For integrals of rational functions, I know that I need to split into fractions, where the polynomials at the numerator are maximum second order polynomials in $x$ or are the type $x^n$. Then I complete the square. If I get something like $$\int\frac{ax+b}{(ax+b)^2+c^2}dx $$ then I can change variables and get $\ln$. If I get $$\int\frac 1{1+x^2}dx$$ then I know that it's $\arctan$. Everything else I can derive

Answer 2

The pythagorean theorem, as applied to trigonometry says

$\sin^2 \theta + \cos^2 \theta = 1$

This is the key piece of knowledge for these integrals.

The implications are:

$\cos \theta = \pm \sqrt {1-\sin^2 \theta}\\ \sin \theta = \pm \sqrt {1-\cos^2 \theta}\\ \tan^2 \theta + 1 = \sec^2 \theta$

How does this relate to these integrals...

Whenever you see $x^2 + 1$ in someplace inconvenient, like under a radical or in the denominator, you should be thinking of the substitution $x = \tan \theta.$ With this substitution it will become $\tan^2\theta + 1 = \sec^2 \theta$

Similarly, when you see $1-x^2$ you should be thinking $x=\sin\theta$ or $x = \cos \theta$ and the expression becomes $1-\sin^2\theta = \cos^2\theta$

And when you see $x^2 - 1$ it is a bit of a toss up. Sometimes, $x = \sin \theta$ works and sometimes $x = \sec\theta$ works better. It really has to do whether you have reason to think $|x|<1$ (in which case use the sine substitution) or $|x| > 1$ in which case use the secant substitution.

Taking it up a level.

When you see $x^2 + a^2$ then you should be thinking $x = a\tan \theta$ and when you see $a^2 x^2 + b^2$ think $x = \frac {b}{a}\tan \theta$ Finally, when you see $(x+a)^2 + b^2,$ think $x+a = b\tan \theta.$ These will simplify nicely.

Some examples. The area of a portion of a circle...

The equation of our circle is $x^2 + y^2 = 1$

We want $\int_a^1 \sqrt {1-x^2} \ dx$

start with: $x = \cos \theta\\ dx = -\sin\theta\ d\theta$

What happens to our limits of integration?

$a = \cos \theta\\ \theta = \arccos a\\ 1 = \cos \theta\\ \theta = 0$

$\int_{\arccos a}^{0} \sqrt {1-\cos^2\theta} (-\sin\theta \ d\theta)$

We can reverse the order of integration if we change the sign. $1-\cos^2 \theta = \sin^2\theta$

$\int_0^{\arccos a} \sqrt {\sin^2\theta} (\sin\theta) \ d\theta\\ \int_0^{\arccos a} \sin^2\theta \ d\theta$

Apply a half-angle identity:

$\sin^2\theta = \frac 12 (1-\cos 2\theta)$

$\int_0^{\arccos a} \frac 12 (1-\cos 2\theta) \ d\theta$

$\frac 12 (\theta-\frac 12 \sin 2\theta)|_0^{\arccos a}$

At this point I like to use the double angle identity

$\frac 12 (\theta-\sin \theta\cos \theta)|_0^{\arccos a}$

$\sin \arccos a = \sqrt {1-a^2}$

$\frac 12 (\arccos a - a\sqrt {1-a^2})$

What does this mean geometrically?

The area of the red plus the green is $\frac 12 \theta = \frac 12 \arccos a$

The height of the red triangle is $\sqrt {1-a^2}$ and the area is $\frac 12 a\sqrt {1-a^2}$

One more example

$\int \frac {1}{x^2+x+1} \ dx$

The denominator looks like a bit of a bear. It doesn't factor, if it did, I would suggest partial fractions. As it doesn't we use "completing the square."

$x^2 + x + 1 = (x+\frac 12)^2 + \frac 34$

$\int \frac {1}{(x+\frac 12)^2 + \frac 34} \ dx$

$x+\frac 12 = \sqrt {\frac 34} \tan \theta\\ dx = \sqrt {\frac 34} \sec^2 \theta\ d\theta$

Don't let those radicals scare you, they are just constants.

$\int \frac {\sqrt {\frac 34}\sec^2\theta}{\frac 34 \tan^2\theta + \frac 34} \ d\theta\\ \frac {1}{\sqrt {\frac 34}}\int \frac {\sec^2\theta}{\sec^2\theta} \ d\theta\\ \frac {2}{\sqrt 3} \theta $

Now we need to reverse the substitution

$x + \frac 12 = \sqrt {\frac 34} \tan \theta\\ \frac {2}{\sqrt 3} (x+\frac 12) = \tan \theta\\ \theta = \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$

$\frac {2\sqrt 3}{3} \arctan (\frac {2\sqrt 3}{3}x + \frac {\sqrt 3}{3})$

I hope this helps.