Week 9 Discussion Notes

Table of Contents

• Functions

• Inverses

  • What is an Inverse?
  • When is there an Inverse?

• Inverse Trigonometric Functions

  • Definitions
  • Derivatives

• Integration by Parts

  • Examples

Functions

A function $f$ always comes with a domain $D$, and the range of $f$ is determined by $D$. If the domain isn't specified, then it's usually taken to be the largest set where $f$ is defined.

Example 1.

If $f\p{x} = \frac{1}{x}$ and $D$ is not specified, then we assume $D = \set{x \in \R \mid x \neq 0}$.

Example 2.

Let $f\p{x} = x^2$.

• $D = \R \implies R = \left[ 0, \infty \right)$

• $D = \left[ 0, \infty \right) \implies R = \left[ 0, \infty \right)$

• $D = \set{x \in \R \mid x > 2} \implies R = \p{4, \infty}$

Inverses

What is an Inverse?

Given a function $f$, we'd like to know if we can go "backwards" from $f$. This is useful if you need to solve something like $f\p{x} = y$; if we can "undo" $f$, then we automatically have the solution to the equation. This can be done if $f$ is invertible:

Definition

Let $f$ have domain $D$ and range $R$. If $\exists g$ with domain $R$ and range $D$ such that

• $f\p{g\p{x}} = x$ for all $x \in R$ and
• $g\p{f\p{x}} = x$ for all $x \in D$,

then we say $f$ is invertible and we call $f^{-1} = g$ the inverse of $f$.

When is there an Inverse?

So when does $f$ have an inverse? Let's look at a situation where $f$ does not have one.

$f$ doesn't have an inverse here because we can't go backwards from $y$ to get both $a$ and $b$. Indeed, any function $g$ with domain $R$ will map $y$ to either $a$ or $b$, but not both. The problem is that $f$ maps two different elements to the same thing, so if we prevent this, $f$ will have an inverse.

To prevent this, we need $f$ to map two different elements to two different things. More succinctly, $a \neq b \implies f\p{a} \neq f\p{b}$. If $f$ satisfies this, then we call $f$ one-to-one, but this isn't the only way to tell if $f$ is one-to-one:

Definition (one-to-one)

Let $f$ have domain $D$, and suppose one of the following is true about $f$:

• $a \neq b \implies f\p{a} \neq f\p{b}$ for any $a, b \in D$ or
• $f\p{x} = y$ has exactly one solution for every $y \in R$ or
• $f\p{a} = f\p{b} \implies a = b$ for any $a, b \in D$ or
• $f$ passes the horizontal line test.

Then we say $f$ is one-to-one or injective.

Notice that the domain is very important when determining whether $f$ is one-to-one or not.

Example 3.

Let $f\p{x} = x^2$.

• If $D = \R$, then $f$ is not one-to-one.

  $f\p{x} = 1$ has two solutions: $-1$ and $1$.

• If $D = \left[ 0, \infty \right)$, then $f$ is one-to-one.

  Since $f$ is one-to-one, it has an inverse, which is $f^{-1}\p{x} = \sqrt{x}$.

Inverse Trigonometric Functions

Definitions

In order to define inverse trig functions, we need to find domains so that they are one-to-one without changing their ranges. Because trig functions oscillate, there are a lot of intervals like these, so which one do we take? $0$ is a nice number, so we are going to look for a domain containing $0$ which doesn't change the range of our function.

Take a look at the graph of $\sin{x}$:

As you can see, $\br{-\frac{\pi}{2}, \frac{\pi}{2}}$ does the trick. On this interval, $\sin{x}$ passes the horizontal line test and it hits every point in its range, $\br{-1, 1}$.

Now look at $\cos{x}$:

There are two intervals in this case: $\br{-\pi, 0}$ and $\br{0, \pi}$. Hopefully, you agree that $\br{0, \pi}$ is "nicer" than the other one since it contains only positive numbers, so we're going to restrict $\cos{x}$ to $\br{0, \pi}$ when defining the inverse.

The last one we're going to look at is $\tan{x}$:

As in the case with $\cos{x}$, we have multiple choices, but I think it's clear that $\p{-\frac{\pi}{2}, \frac{\pi}{2}}$ is the "nicest" interval we can use.

We can play the same game with the rest of the trig functions, and what we get are the following inverse functions:

Definition (inverse trigonometric functions)

We define the inverses of trig functions via the following tables:

$$\def\arraystretch{1.5} \begin{array}{lll} f\p{x} & f^{-1}\p{x} & \text{range} \\ \hline \sin{x} & \arcsin{x} & \br{-\frac{\pi}{2}, \frac{\pi}{2}} \\ \csc{x} & \arccsc{x} & \left\lbrack -\frac{\pi}{2}, 0 \right\rparen \cup \left\lparen 0, \frac{\pi}{2} \right\rbrack \\ \tan{x} & \arctan{x} & \p{-\frac{\pi}{2}, \frac{\pi}{2}} \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{lll} f\p{x} & f^{-1}\p{x} & \text{range} \\ \hline \cos{x} & \arccos{x} & \br{0, \pi} \\ \sec{x} & \arcsec{x} & \left\lbrack 0, \frac{\pi}{2} \right\rparen \cup \left\lparen \frac{\pi}{2}, \pi \right\rbrack \\ \cot{x} & \arccot{x} & \p{0, \pi} \end{array}$$

We get the range of $f^{-1}\p{x}$ by swapping the domain and range of $f\p{x}$.

Exercise 1.

Figure out the domains of the inverse trig functions.

One thing to notice is how I arranged the tables. If you look at the top table, the ranges are all $\br{-\frac{\pi}{2}, \frac{\pi}{2}}$ with some points removed, i.e., the range of $\arcsin{x}$ with some points removed. Similarly, the bottom table ranges are all the range of $\arccos{x}$ with some points removed.

The way I remember how to group them are as follows: if $\sin{x}$ is the "main" function (e.g., $\csc{x} = \frac{1}{\sin{x}}$), then the range is related to $\arcsin{x}$. Similarly, if $\cos{x}$ is the "main" function, then the range will be related to $\arccos{x}$.

Derivatives

Now that we have the inverses, let's talk about their derivatives:

Theorem

$$\begin{aligned} &\deriv{}{x} \arcsin{x} &=& \phantom{-}\frac{1}{\sqrt{1 - x^2}} \\ &\deriv{}{x} \arccos{x} &=& -\frac{1}{\sqrt{1 - x^2}} \\ &\deriv{}{x} \arctan{x} &=& \phantom{-}\frac{1}{1 + x^2} \\ &\deriv{}{x} \arcsec{x} &=& \phantom{-}\frac{1}{\abs{x}\sqrt{x^2 - 1}} \\ &\deriv{}{x} \arccsc{x} &=& -\frac{1}{\abs{x}\sqrt{x^2 - 1}} \\ &\deriv{}{x} \arccot{x} &=& -\frac{1}{1 + x^2} \\ \end{aligned}$$

These are all done by implicit differentiation, and I'll show one of them:

Example 4.

Prove that $\deriv{}{x} \arcsec{x} = \frac{1}{\abs{x}\sqrt{x^2 - 1}}$.

Solution.

To start, I'm going to set $y = \arcsec{x}$. Since $\arcsec{x}$ and $\sec{x}$ are inverses, we can apply $\sec{\theta}$ to both sides to get $\sec{y} = x$. Now we can take derivatives on both sides to get

$$\p{\sec{y}\tan{y}} y' = 1 \implies y' = \frac{1}{\sec{y}\tan{y}}.$$

To finish the problem, we need to get the right-hand side in terms of $x$. We already know that $\sec{y} = x$, so we need to handle $\tan{y}$. We can use the following trig identity to relate $\tan{y}$ to $\sec{y} = x$:

$$\begin{aligned} \tan^2{y} + 1 = \sec^2{y} = x^2 &\implies \tan^2{y} = x^2 - 1 \\ &\implies \tan{y} = \pm\sqrt{x^2 - 1}. \end{aligned}$$

We need to figure out when the sign is positive and when the sign is negative, so we need to break it into two cases:

Case 1: $\sec{y} = x > 0$

Since $y = \arcsec{x}$, $y$ is in the range of $\arcsec{x}$, so $y$ is in the first or second quadrant.

Since $\sec{y} > 0$, this means that $\cos{y} > 0$, which means that $y$ must be in Quadrant I. In this quadrant, $\tan{y} > 0$, so $\tan{y} = \sqrt{x^2 - 1}$ in this case.

Case 2: $\sec{y} = x < 0$

We can use the same argument: in this case, $\cos{y} < 0$, so $y$ is in Quadrant II, which means that $\tan{y} < 0$. Thus, $\tan{y} = -\sqrt{x^2 - 1}$ in this case.

Now that we have established the signs of $\tan{y}$, we just need to plug it back into our original function. There are two cases, so we end up with a piecewise function:

$$\begin{aligned} y' &= \frac{1}{x\tan{y}} \\ &= \begin{cases} \dfrac{1}{\phantom{-}x\sqrt{x^2 - 1}} & \text{if } x > 0 \\[2ex] \dfrac{1}{-x\sqrt{x^2 - 1}} & \text{if } x < 0 \end{cases} \\ &= \frac{1}{\abs{x}\sqrt{x^2 - 1}}. \end{aligned}$$

The last equality comes from the fact that $\abs{x} = x$ if $x > 0$ and $-x$ if $x < 0$.

Integration by Parts

Theorem (integration by parts)

If $u$ and $v$ are functions, then

$$\int u \,\diff{v} = uv - \int v \,\diff{u}.$$

Given an integral, you're free to choose $u$ and $\diff{v}$ however you like. However, the main issue after applying the formula is the integral term $\int v \,\diff{u}$, so before you start, try to think about what $v \,\diff{u}$ looks like.

The way I remember the formula is that you pick one thing to differentiate ($u$) and one thing to integrate ($\diff{v}$). Then you integrate first to get $uv$ and then differentiate after to get $v \,\diff{u}$.

Examples

Example 5.

Calculate $\displaystyle \int x\sin{x} \,\diff{x}$.

Solution.

For this problem, you'll want to differentiate $x$ because it will just become $1$. If you integrate $x \,\diff{x}$, then you end up with $\frac{1}{2} x^2$, which is more complicated. So, we'll use $u = x$ and $\diff{v} = \sin{x} \,\diff{x}$, which gives us

$$\diff{u} = \diff{x} \quad\text{and}\quad v = -\cos{x}.$$

Thus,

$$\begin{aligned} \int x\sin{x} \,\diff{x} &= -x\cos{x} - \int -\cos{x} \,\diff{x} \\ &= -x\cos{x} + \int \cos{x} \,\diff{x} \\ &= \boxed{-x\cos{x} + \sin{x} + C}. \end{aligned}$$

Example 6.

Calculate $\displaystyle \int \cos^{-1}{x} \,\diff{x}$.

Solution.

Even though there's only one term, we can still use integration by parts because $\cos^{-1}{x} = \cos^{-1}{x} \cdot 1$. Choose $\diff{v} = \cos^{-1}{x} \,\diff{x}$ isn't going to do anything for us because we don't know how to integrate it (yet), so we need to let $u = \cos^{-1}{x}$ and $\diff{v} = \diff{x}$ instead. Then

$$\diff{u} = -\frac{\diff{x}}{\sqrt{1 - x^2}} \quad\text{and}\quad v = x.$$

So if we integrate by parts, we get

$$\begin{aligned} \int \cos^{-1}{x} \,\diff{x} &= x\cos^{-1}{x} - \int -\frac{x}{\sqrt{1 - x^2}} \,\diff{x} \\ &= x\cos^{-1}{x} + \int \frac{x}{\sqrt{1 - x^2}} \,\diff{x} \\ &= \boxed{x\cos^{-1}{x} - \sqrt{1 - x^2} + C}. \end{aligned}$$

(To calculate the integral, you can use $u = 1 - x^2$.)

Example 7.

Calculate $\displaystyle \int \ln{x} \,\diff{x}$.

Solution.

Like the example before, we can let $u = \ln{x}$ and $\diff{v} = \diff{x}$. Then

$$\diff{u} = \frac{\diff{x}}{x} \quad\text{and}\quad v = x,$$

so we get

$$\begin{aligned} \int \ln{x} \,\diff{x} &= x\ln{x} - \int x \cdot \frac{1}{x} \,\diff{x} \\ &= x\ln{x} - \int \,\diff{x} \\ &= \boxed{x\ln{x} - x + C}. \end{aligned}$$

Example 8.

Calculate $\displaystyle \int \frac{\ln{x}}{x^3} \,\diff{x}$.

Solution.

I want to differentiate $\ln{x}$ because it gives me $\frac{1}{x}$, which plays nicely with all the other terms. So, we want to choose $u = \ln{x}$ and $\diff{v} = \frac{1}{x^3} \,\diff{x}$, which gives

$$\diff{u} = \frac{\diff{x}}{x} \quad\text{and}\quad v = -\frac{1}{2x^2}.$$

Then the integral becomes

$$\begin{aligned} \int \frac{\ln{x}}{x^3} \,\diff{x} &= -\frac{\ln{x}}{2x^2} - \int -\frac{\diff{x}}{2x^3} \\ &= -\frac{\ln{x}}{2x^2} + \int \frac{\diff{x}}{2x^3} \\ &= -\frac{\ln{x}}{2x^2} - \frac{1}{4x^2} + C \\ &= \boxed{-\p{\frac{2\ln{x} - 1}{4x^2}} + C}. \end{aligned}$$