Integration Techniques

This chapter assumes that you know integrals and sine and cosine derivatives. It also uses weird trig notation

Nowadays it's easiest to use sympy for doing difficult integrals, but in this chapter we'll look into how integrals were done before computers and calculators. It sounds very boring and hard, but you might find especially u-substitution kind of amazing.

U Substitution

In the integral chapter we learned that $\int f(x)\ dx$ means the antiderivatives of $f$. The notation with $dx$'s in it is known as Leibniz's notation, and I explained the $\int f(x)\ dx$ like Leibniz did. However, the notation probably wouldn't be used today if the trick I'm about to show wouldn't make it really handy.

Let's say that $F$ is an antiderivative of $f$; that is, $F'(x)=f(x)$. In the integral chapter we turned the chain rule into an integral rule by integrating both sides:

$$\begin{align}f(g(x))g'(x) &= \frac{d}{dx}F(g(x)) \\ \int f(g(x))g'(x)\ dx &= F(g(x)) + C\end{align}$$

Something interesting happens if we define another variable $u=g(x)$. Then $\frac{du}{dx}=g'(x)$. You can use whatever variable name you want, but $u$ is traditionally used here.

$$\begin{align}\int f(u)\frac{du}{dx}\ dx &= F(u) + C\end{align}$$

Could we just cancel out the $dx$'s? It feels kind of wrong because the $dx$ is just a part of our antiderivative notation and not really a variable, but let's try it and see what happens:

$$\begin{align}\int f(u)\frac{du}{\rcancel{dx}}\rcancel{dx} &\mathop{=}^{\text{wat}} F(u) + C \\ \int f(u)\ du\ &\mathop{=}^{\text{lol}}_{\text{srsly?}}\ F(u) + C\end{align}$$

Amazing, it really worked! $F(u) + C$ is indeed the integral of $f(u)$ with respect to $u$.

$$\begin{align}\frac{d}{du}(F(u)+C) \mathop{=}^{\text{yes}} F'(u)+0 \mathop{=}_{\text{srsly}} f(u)\end{align}$$

We can get to the same result like this:

$$\begin{align}& \int f(\green{g(x)})\blue{g'(x)\ dx} \qquad\qquad \begin{array}{l} \green u = \green{g(x)} \\ \frac{du}{dx} = g'(x) \\ \blue{du} \overset{\text{lol}}= \blue{g'(x)\ dx} \\ \end{array} \\ =&\ \int f(\green u)\ \blue{du} \\ =&\ F(u) + C \\ =&\ F(g(x)) + C\end{align}$$
<__Myst__>  Akuli: that's magic
<Akuli>     yes it is :D
<Akuli>     well not really
<Akuli>     you can kind of understand it if you think of du and dx as
            very small changes of u and x
<Akuli>     and ∫f(u)du as a sum of f(u) by du area rectangles
<Akuli>     that's obviously not a very modern way to define derivatives and
            integrals, but it kind of explain-ishs it

When our teacher had shown this trick to us for the first time, everyone were just amazed by it, and the teacher said that he wanted to take a picture of our faces.

Example: $\int x\sin(x^2)\ dx$

The derivative of $x^2$ is $2x$ and it differs from $x$ by just a constant, so a substitution like $u=x^2$ will work.

$$\begin{align}\int x\sin(x^2)\ dx &= \int \sin(\green{x^2})\ \blue{x\ dx} \qquad\qquad \begin{array}{l} \green u = \green{x^2} \\ \frac{du}{dx} = 2x \\ du = 2x\ dx \\ \blue{x\ dx} = \blue{\frac 1 2 du} \\ \end{array} \\ &= \int\sin(\green u)\blue{\frac 1 2 du} \\ &= \frac 1 2 \int\sin(u)\ du \\ &= \frac 1 2 (-\cos(\green u)) + C \\ &= -\frac 1 2 \cos(\green{x^2}) + C\end{align}$$$$\begin{align}\frac{d}{dx}\left(-\frac 1 2 \cos(x^2) + C\right) = -\frac 1 2 (-\sin(x^2)) \cdot 2x + 0 = x\sin(x^2)\end{align}$$

This worked because it's easy to change the integral so that it contains the derivative of the inside:

$$\begin{align}\int x\sin(x^2)\ dx = \frac 1 2 \int 2x\sin(x^2)\ dx\end{align}$$

Treating $du$ and $dx$ as variables like $du=2x\ dx$ is hand-wavy, but it's very commonly used and accepted hand-waviness because it works so nicely. In fact, many people write $\int dx$ when they mean $\int 1\ dx$ and $\int \frac{dx}{f(x)}$ when they mean $\int \frac{1}{f(x)}\ dx$.

Note that in the above example we added another variable, and switched everything to use that variable instead of $x$; there must be no $x$ left over. If there is, try another $u$ choice, do something to the function before integrating it or maybe try some other integration technique instead of $u$ substitution. When we were done, we went from the $u$ world back to the $x$ world.


Here's a true story. At school we had trouble with this integral:

$$\begin{align}\int \frac{\cos(x)}{\sin^2(x)}\ dx\end{align}$$

Our supply teacher came up with a solution, and when I saw it I knew it was very incorrect, aka WRONG in CAPS AND BOLD. It started like this:

$$\begin{align}\int \frac{\cos(x)}{\sin^2(x)}\ dx = \frac{1}{2\sin(x)}\int\frac{2\sin(x)\cos(x)}{\sin^2(x)}\ dx =\ ...\end{align}$$

Here the supply teacher wanted to have the derivative of $\sin^2(x)$ on the top. You should know by now that we cannot bring non-constants like $\frac{1}{2\sin(x)}$ out of integrals. I argued with the teacher about his solution for the rest of the lesson.

<Zaab1t>    and then everyone clapped
<Akuli>     well, everyone laughed several times
<Akuli>     "this is constant on this interval balblabla"  "NO its a
            function, not a constant"

The following day our actual teacher thanked me for arguing about it, and later I found two more fatal mistakes from the solution.

Anyway, integrate $\dfrac{\cos(x)}{\sin^2(x)}$ correctly. It's easy with the correct substitution. (Note that $x^{-a} = 1/x^a$.)

Let's see how this works for definite integral; that is, $\int_a^b$ instead of $\int$, area instead of antiderivative.

$$\begin{align}\int_a^b f(g(x))g'(x)\ dx &= \bigl[F(g(x))\bigr]_{x=a}^{x=b} \\ &= F(\green{g(b)})-F(\green{g(a)}) \\ &= \bigl[F(\green{u})\bigr]_{u=\green{g(a)}}^{u=\green{g(b)}} \\ &= \int_{g(a)}^{g(b)} f(u)\ du\end{align}$$

This means that we just need to change the limits to $u$ values.

Example: $\int_\sqrt{\tau}^\sqrt{2\tau} x\sin(x^2)\ dx$

$$\begin{align}\int_\sqrt{\tau}^\sqrt{2\tau} x\sin(x^2)\ dx &= \int_\sqrt{\tau}^\sqrt{2\tau} \sin(x^2)\ x\ dx \qquad\qquad \begin{array}{l} u = x^2 \\ \text{if }x=\sqrt{\tau}\text{ then }u=\tau \\ \text{if }x=\sqrt{2\tau}\text{ then }u=2\tau \\ du = 2x\ dx \\ x\ dx = \frac 1 2 du \\ \end{array} \\ &= \int_\tau^{2\tau}\sin(u)\frac 1 2 du \\ &= \frac 1 2 \bigl[-\cos(u)\bigr]_{u=\tau}^{u=2\tau} \\ &= \frac 1 2 (-\cos(2\tau)-(-\cos(\tau))) \\ &= \frac 1 2 (\green{\cos(\tau)-\cos(2\tau)})\end{align}$$

$\tau$ and $2\tau$ differ from each other by a full turn, and rotating a full turn on the unit circle brings us back to where we started and the cosines are the same.

$$\begin{align}\cos(\tau) &= \cos(2\tau) \\ \green{\cos(\tau)-\cos(2\tau)} &= 0 \\ \int_\sqrt{\tau}^\sqrt{2\tau} x\sin(x^2)\ dx &= 0\end{align}$$

Note that we didn't need to go back to the $x$ world; with definite integrals it's enough to change the limits, and then we can forget the whole $x$.


The parabola $y=(x-1)^2$ rotates around the $x$ axis between $0 \le x \le 2$. Calculate the volume of the shape it forms.

Trig Substitution

Let's say we have an integral like this:

$$\begin{align}\int\sqrt{1-x^2}\ dx\end{align}$$

A substitution like $u=1-x^2$ does not work here because the derivative of the inside is $-2x$. The $-2$ is not a problem, but we have no extra $x$ in the integral (and obviously we can't just add an $x$ there).

Here we proved this formula as an exercise:

$$\begin{align}\sin^2(\theta) + \cos^2(\theta) = 1\end{align}$$

We can solve $\cos^2(\theta) = 1-\sin^2(\theta)$. So, wouldn't it be handy if we had $x=\sin(\theta)$?

$$\begin{align}\newcommand{abs}[1]{\lvert{#1}\rvert} \\ \sqrt{1-(\sin(\theta))^2} = \sqrt{\cos^2(\theta)} = \abs{\cos(\theta)}\end{align}$$

We can do this with a substitution $\theta = \arcsin(x)$. Here $\arcsin$ is the inverse of $\sin$, but it always returns values so that $-\frac \tau 4 \le \theta \le \frac \tau 4$, denoted with blue in this pic:

This is because if we know what $\sin(\theta)$ is, there are multiple values that $\theta$ could be, like the red dot in the image, but we don't have this problem on a restricted range like $-\frac \tau 4 \le \theta \le \frac \tau 4$.

Anyway, you can see from the pic that the $x$ coordinates are not negative, so we have $\cos(\theta) \ge 0$. Let's do the integral

$$\begin{align}&\int\sqrt{1-x^2}\ dx \qquad\qquad \begin{array}{l} \\ \theta=\arcsin(x) \\ x=\sin(\theta) \\ dx = \cos(\theta)\ d\theta \\ \green{\cos(\theta)} \ge 0 \\ \end{array} \\ =&\ \int\sqrt{1-\sin^2(\theta)}\ \cos(\theta)\ d\theta \\ =&\ \int\sqrt{\cos^2(\theta)}\ \cos(\theta)\ d\theta \\ =&\ \int\abs{\green{\cos(\theta)}}\cos(\theta)\ d\theta \\ =&\ \int\cos^2(\theta)\ d\theta\end{align}$$

We still can't use u substitution because we don't have the derivative of $\cos(\theta)$ anywhere. Let's cheat and use Wikipedia's trig formula list:

$$\begin{align}\cos^2(\theta) = \frac{1+\cos(2\theta)}{2}\end{align}$$

We'll learn an easy way to prove formulas like this in the Euler formula chapter.

$$\begin{align}\int\sqrt{1-x^2}\ dx = \int\frac{1+\cos(2\theta)}{2}\ d\theta = \frac 1 2 \int(1+\cos(2\theta))\ d\theta\end{align}$$

Rest of this is quite straight-forward, so I don't see any reason to go through that.

Here's a handy-dandy table of trig substitutions with a positive constant $a$:

If your integral
contains this...
...use this sub... ...and this trig formula... ...that comes from here
$\sqrt{a-x^2}$ $x=\sqrt a \sin(\theta)$ $\sin^2(\theta)+\cos^2(\theta)=1$ this exercise
$\sqrt{a+x^2}$ $x=\sqrt a \tan(\theta)$ $1+\tan^2(\theta)=\frac{1}{\cos^2(\theta)}$ the tan derivative


In the integral chapter I showed you this pic of different ways to prove the circle area $\pi r^2$:

Prove the circle area formula using the first way, the one with the stupid vertical slices.