# Trigonometric Integrals

Let us attempt to calculate $\int\cos^n xdx$ where $n$ is a positive integer. In the following table, the first column represents $\cos^{n-1}x$ and its derivative, and the second column represents $\cos x$ and its integral.
$$\begin{array}{ccc} \cos^{n-1}x & & \cos x\\ &\stackrel{+}{\searrow}&\\ -(n-1)\cos^{n-2}x\sin x & \stackrel{-}{\longrightarrow} & \sin x\\ \end{array}$$
By integration by parts, we have
\begin{align*}
\int\cos^n xdx&=\cos^{n-1}x\sin x+(n-1)\int\cos^{n-2}x\sin^2xdx\\
&=\cos^{n-1}x\sin x+(n-1)\int\cos^{n-2}xdx-(n-1)\int\cos^{n-1}xdx+C’
\end{align*}
where $C’$ is a constant. Solving this for $\int\cos^nxdx$, we obtain

\label{eq:cosred}
\int\cos^n xdx=\frac{1}{n}\cos^{n-1}x\sin x+\frac{n-1}{n}\int\cos^{n-2}xdx+C

where $C=\frac{C’}{n}$. The formula such as \eqref{eq:cosred} is called a reduction formula. Similarly we obtain the following reduction formulae.
\begin{align}
\int\sin^n xdx&=-\frac{1}{n}\sin^{n-1}x\cos x+\frac{n-1}{n}\int\sin^{n-2}dx\\
\int\tan^nxdx&=\frac{1}{n-1}\tan^{n-1}x-\int\tan^{n-2}dx,\ n\ne 1\\
\int\sec^nxdx&=\frac{1}{n-1}\sec^{n-2}x\tan x+\frac{n-2}{n-1}\int\sec^{n-2}xdx,\ n\ne 1
\end{align}
Example. Use the reduction formula \eqref{eq:cosred} to evaluate $\int\cos^3xdx$.

Solution.
\begin{align*}
\int\cos^3xdx&=\frac{1}{3}\cos^2x\sin x+\frac{2}{3}\int\cos xdx\\
&=\frac{1}{3}\cos^2x\sin x+\frac{2}{3}\sin x+C,
\end{align*}
where $C$ is a constant.

Integral like the following example is rather tricky.

Example. Evaluate $\int\sec xdx$.

Solution.
\begin{align*}
\int\sec xdx&=\int\sec x\frac{\sec x+\tan x}{\sec x+\tan x}dx\\
&=\int\frac{\sec^2x+\sec x\tan x}{\sec x+\tan x}dx\\
&=\frac{du}{u}\ (\mbox{substitution}\ u=\sec+\tan x)\\
&=\ln|u|+C\\
&=\ln|\sec x+\tan x|+C,
\end{align*}
where $C$ is a constant.

Example. Evaluate $\int\csc xdx$.

Solution. It can be done similarly to the previous example.
\begin{align*}
\int\csc xdx&=\int\csc x\frac{\csc x+\cot x}{\csc x+\cot x}dx\\
&=-\ln|\csc x+\cot x|+C,
\end{align*}
where $C$ is a constant.

Evaluating Integrals of the Type $\int\sin^mx\cos^nxdx$ Where $m,n$ Are Positive Integers

Case 1. One of the integer powers, say $m$, is odd.

$m=2k+1$ for some integer $k$. So,
\begin{align*}
\sin^mx&=\sin^{2k+1}x\\
&=(\sin^2x)^k\sin x\\
&=(1-\cos^2x)^k\sin x.
\end{align*}
Use the substitution $u=\cos x$ in this case.

Example. Evaluate $\int\sin^3x\cos^2xdx$.

Solution.
\begin{align*}
\int\sin^3x\cos^2xdx&=\int \sin^2x\sin x\cos^2xdx\\
&=\int(1-\cos^2x)\cos^2x\sin xdx\\
&=-\int(1-u^2)u^2du\ (\mbox{substition}\ u=\cos x)\\
&=\frac{u^5}{4}-\frac{u^3}{3}+C\\
&=\frac{\cos^5x}{5}-\frac{\cos^3x}{3}+C,
\end{align*}
where $C$ is a constant.

Example. Evaluate $\int\cos^3xdx$.

Solution.
\begin{align*}
\int\cos^3xdx&=\int\cos^2x\cos xdx\\
&=\int(1-\sin^2x)\cos xdx\\
&=\int(1-u^2)du\ (\mbox{substitution}\ u=\sin x)\\
&=u-\frac{u^3}{3}+C\\
&=\sin x-\frac{\sin^3x}{3}+C,
\end{align*}
where $C$ is a constant.

Case 2. If both $m$ and $n$ are even.

In this case, use the trigonometric identities
$$\sin^2x=\frac{1-\cos 2x}{2},\ \cos^2x=\frac{1+\cos 2x}{2}.$$

Example. Evaluate $\int\sin^2x\cos^4xdx$.

$$\frac{1}{16}\left(x-\frac{1}{4}\sin 4x+\frac{1}{3}\sin^32x\right)+C,$$
where $C$ is a constant.

Integrals of Powers of $\tan x$ and $\sec x$

This type of integrals can be mostly done by using the trigonometric identity
$$1+\tan^2x=\sec^2x.$$

Example. Evaluate $\int\tan^4xdx$.

Solution.
\begin{align*}
\int\tan^4xdx&=\int\tan^2x\tan^2xdx\\
&=\int\tan^2x(\sec^2x-1)dx\\
&=\int\tan^2x\sec^2xdx-\int\tan^2xdx\\
&=\int u^2du-\int(\sec^2x-1)dx\ (\mbox{substition}\ u=\tan x)\\
&=\frac{\tan^3x}{3}-\tan x+x+C,
\end{align*}
where $C$ is a constant.

Example. Evaluate $\int\sec^3xdx$.

Solution.
\begin{align*}
\int\sec^3xdx&=\int\sec x\sec^2xdx\\
&=\sec x\tan x-\int\tan^2x\sec xdx\ (\mbox{integration by parts})\\
&=\sec x\tan x-\int(\sec^2x-1)\sec xdx\\
&=\sec x\tan x-\int\sec^3xdx+\int\sec xdx+C’\\
&=\sec x\tan x-\int\sec^3xdx+\ln|\sec x+\tan x|+C’,
\end{align*}
where $C’$ is a constant. Hence,
$$\int\sec^3xdx=\frac{1}{2}\sec x\tan x+\frac{1}{2}\ln|\sec x+\tan x|+C,$$
where $C=\frac{C’}{2}$.

Products of Sines and Cosines

This type of integrals include $\int\sin mx\cos nxdx$, $\int\sin mx\cos nxdx$, and $\int\cos mx\cos nxdx$. In this case use the identities
\begin{align*}
\sin mx\sin nx&=\frac{1}{2}[\cos(m-n)x-\cos(m+n)x]\\
\sin mx\cos nx&=\frac{1}{2}[\sin(m-n)x+\sin(m+n)x]\\
\cos mx\cos nx&=\frac{1}{2}[\cos(m-n)x+\cos(m+n)x]
\end{align*}
Example. Evaluate $\int\sin 3x\cos5xdx$.

Solution.
\begin{align*}
\int\sin 3x\cos5xdx&=-\frac{1}{2}\int\sin 2x+\frac{1}{2}\int\sin 8xdx\\
&=\frac{1}{4}\cos 2x-\frac{1}{16}\cos 8x+C,
\end{align*}
where $C$ is a constant.

Example. Evaluate $\int_0^1\sin m\pi x\sin n\pi xdx$ and $\int_0^1\cos m\pi x\cos n\pi xdx$ where $m$ and $n$ are positive integers.

Solution. If $m=n$, then
\begin{align*}
\int_0^1\sin m\pi x\sin n\pi xdx&=\int_0^1\sin^2m\pi xdx\\
&=\int_0^1\frac{1-\cos 4m\pi x}{2}dx\\
&=\frac{1}{2}\int_0^1dx-\frac{1}{2}\int_0^1\cos 4m\pi xdx\\
&=\frac{1}{2}-\frac{1}{8m\pi}[\sin 4m\pi x]_0^1\\
&=\frac{1}{2}.
\end{align*}
Now we assume that $m\ne n$. Then
\begin{align*}
\int_0^1\sin m\pi x\sin n\pi xdx&=\frac{1}{2}\int_0^1\cos(m-n)\pi xdx-\frac{1}{2}\int_0^1\cos(m+n)\pi xdx\\
&=0.
\end{align*}
So, we can simply write the integral as

\label{eq:orthofunct}
\int_0^1\sin m\pi x\sin n\pi xdx=\frac{1}{2}\delta_{mn},

where
$$\delta_{mn}=\left\{\begin{array}{ccc} 1 & \mbox{if} & m=n,\\ 0 & \mbox{if} & m\ne 0. \end{array}\right.$$
$\delta_{mn}$ is called the Kronecker’s delta.
Similarly, we also have

\label{eq:orthofunct2}
\int_0^1\cos m\pi x\cos n\pi xdx=\frac{1}{2}\delta_{mn}.

The integrals \eqref{eq:orthofunct} and \eqref{eq:orthofunct2} play an important role in studying the boundary value problems with heat equation and wave equation. They also appear in the study of different branches of mathematics and physics such as functional analysis, Fourier analysis, electromagnetism, and quantum mechanics, etc. In mathematics and physics, often functions like $\sin n\pi x$ and $\cos n\pi x$ are treated as vectors and integrals like \eqref{eq:orthofunct}  and \eqref{eq:orthofunct2} can be considered as inner products $\langle\sin m\pi x,\sin n\pi x\rangle$ and $\langle\cos m\pi x,\cos n\pi x\rangle$, respectively. In this sense, we can say that $\sin m\pi x$ and $\sin n\pi x$ are orthogonal if $m\ne n$. For this reason, functions $\sin n\pi x$ and $\cos n\pi x$, $n=1,2,\cdots$ are called orthogonal functions.

# Integration by Parts

Let $f(x)$ and $g(x)$ be differentiable functions. Then the product rule
$$(f(x)g(x))’=f’(x)g(x)+f(x)g’(x)$$

\label{eq:intpart}
\int f(x)g’(x)dx=f(x)g(x)-\int f’(x)g(x)dx.

The formula \eqref{eq:intpart} is called integration by parts. If we set $u=f(x)$ and $v=g(x)$, then \eqref{eq:intpart} can be also written as

\label{eq:intpart2}
\int udv=uv-\int vdu.

Example. Evaluate $\int x\cos xdx$.

Solution. Let $u=x$ and $dv=\cos xdx$. Then $du=dx$ and $v=\sin x$. So,
\begin{align*}
\int x\cos xdx&=x\sin x-\int\sin xdx\\
&=x\sin x+\cos x+C,
\end{align*}
where $C$ is a constant.

Example. Evaluate $\int\ln xdx$.

Solution. Let $u=\ln x$ and $dv=dx$. Then $du=\frac{1}{x}dx$ and $v=x$. So,
\begin{align*}
\int\ln xdx&=x\ln x-\int x\cdot\frac{1}{x}dx\\
&=x\ln x-x+C,
\end{align*}
where $C$ is a constant.

Often it is required to apply integration by parts more than once to evaluate a given integral. In that case, it is convenient to use a table as shown in the following example.

Example. Evaluate $\int x^2e^xdx$

Solution. In the following table, the first column represents $x^2$ and its derivatives, and the second column represents $e^x$ and its integrals.
$$\begin{array}{ccc} x^2 & & e^x\\ &\stackrel{+}{\searrow}&\\ 2x & & e^x\\ &\stackrel{-}{\searrow}&\\ 2 & & e^x\\ &\stackrel{+}{\searrow}&\\ 0 & & e^x. \end{array}$$
This table shows the repeated application of integration by parts. Following the table, the final answer is given by
$$\int x^2e^xdx=x^2e^x-2xe^x+2e^x+C,$$
where $C$ is a constant.

Example. Evaluate $\int x^3\sin xdx$.

Solution. In the following table, the first column represents $x^3$ and its derivatives, and the second column represents $\sin x$ and its integrals.
$$\begin{array}{ccc} x^3 & & \sin x\\ &\stackrel{+}{\searrow}&\\ 3x^2 & & -\cos x\\ &\stackrel{-}{\searrow}&\\ 6x & & -\sin x\\ &\stackrel{+}{\searrow}&\\ 6 & & \cos x\\ &\stackrel{-}{\searrow}&\\ 0 & & \sin x. \end{array}$$
Following the table, the final answer is given by
$$\int x^3\sin xdx=-x^3\cos x+3x^2\sin x+6x\cos x-6\sin x+C,$$
where $C$ is a constant.

Example. Evaluate $\int e^x\cos xdx$.

Solution. In the following table, the first column represents $e^x$ and its derivatives, and the second column represents $\cos x$ and its integrals.
$$\begin{array}{ccc} e^x & & \cos x\\ &\stackrel{+}{\searrow}&\\ e^x & & \sin x\\ &\stackrel{-}{\searrow}&\\ e^x & & -\cos x. \end{array}$$
Now, this is different from the previous two examples. While the first column repeats the same function $e^x$, the functions second column changes from $\cos x$ to $\sin x$ and to $\cos x$ again up to sign. In this case, we stop there and write the answer as we have done in the previous two examples and add to it $\int e^x(-\cos x)dx$. (Notice that the integrand is the product of functions in the last row.) That is,
$$\int e^x\cos xdx=e^x\sin x-e^x\cos x-\int e^x\cos xdx.$$
For now we do not worry about the constant of integration. Solving this for $\int e^x\cos xdx$, we obtain the final answer
$$\int e^x\cos xdx=\frac{1}{2}e^x\sin x-\frac{1}{2}e^x\cos x+C,$$
where $C$ is a constant.

Example. Evaluate $\int e^x\sin xdx$.

Solution. In the following table, the first column represents $e^x$ and its derivatives, and the second column represents $\sin x$ and its integrals.
$$\begin{array}{ccc} e^x & & \sin x\\ &\stackrel{+}{\searrow}&\\ e^x & & -\cos x\\ &\stackrel{-}{\searrow}&\\ e^x & & -\sin x. \end{array}$$
This is similar to the above example. The first columns repeats the same function $e^x$, and the functions in the second column changes from $\sin x$ to $\cos x$ and to $\sin x$ again up to sign. So we stop there and write
$$\int e^x\sin xdx=-e^x\cos x+e^x\sin x-\int e^x\sin xdx.$$
Solving this for $\int e^x\sin xdx$, we obtain
$$\int e^x\sin xdx=-\frac{1}{2}e^x\cos x+\frac{1}{2}e^x\sin x+C,$$
where $C$ is a constant.

Example. Evaluate $\int e^{5x}\cos 8xdx$.

Solution. In the following table, the first column represents $e^{5x}$ and its derivatives, and the second column represents $\cos 8x$ and its integrals.
$$\begin{array}{ccc} e^{5x} & & \cos 8x\\ &\stackrel{+}{\searrow}&\\ 5e^{5x} & & \frac{1}{8}\sin 8x\\ &\stackrel{-}{\searrow}&\\ 25e^{5x} & & -\frac{1}{64}\cos 8x. \end{array}$$
The first columns repeats the same function $e^{5x}$ up to constant multiple, and the functions in the second column changes from $\cos 8x$ to $\sin 8x$ and to $\cos 8x$ again to constant multiple. This case also we do the same.
$$\int e^{5x}\cos 8xdx=\frac{1}{8}e^{5x}\sin 8x+\frac{5}{64}e^{5x}\cos 8x-\frac{25}{64}\int e^{5x}\cos 8xdx.$$
Solving this for $\int e^{5x}\cos 8xdx$, we obtain
$$\int e^{5x}\cos 8xdx=\frac{8}{89}e^{5x}\sin 8x+\frac{5}{89}e^{5x}\cos 8x+C,$$
where $C$ is a constant.

The evaluation of a definite integral by parts can be done as

\label{eq:intpart3}
\int_a^b f(x)g’(x)dx=[f(x)g(x)]_a^b-\int_a^b f’(x)g(x)dx.

Example. Find the area of the region bounded by $y=xe^{-x}$ and the x-axis from $x=0$ to $x=4$.

The graph of y=xexp(-x), x=0..4

Solution. Let $u=x$ and $dv=e^{-x}dx$. Then $du=dx$ and $v=-e^{-x}$. Hence,
\begin{align*}
A&=\int_0^4 xe^{-x}dx\\
&=[-xe^{-x}]0^4+\int_0^4 e^{-x}dx\\
&=-4e^{-4}+[-e^{-x}]_0^4\\
&=1-5e^{-4}.
\end{align*}

# The First and Second Derivative Tests

The First Derivative Test

The derivative $f’(x)$ can tell us a lot about the function $y=f(x)$. It can tell us where critical points are i.e. points at which $f’(x)=0$ and the critical points are likely places at which $y=f(x)$ assumes a local maximum or a local minimum values. By further examining the properties of $f’(x)$ we can also determine at which critical point, $f(x)$ assumes a local maximum, or a local minimum, or neither. But first we see that $f’(x)$ can tell us where $y=f(x)$ is increasing or decreasing.

Theorem. Increasing/Decreasing Test

1. If $f’(x)>0$ on an open interval, $f$ is increasing on that interval.
2. If $f’(x)<0$ on an open interval, $f$ is decreasing on that interval.

Example. Find where $f(x)=3x^4-4x^3-12x^2+5$ is increasing and where it is decreasing.

Solution.
\begin{align*}
f’(x)&=12x^3-12x^2-24x\\
&=12x(x^2-x-2)\\
&=12x(x-2)(x+1).
\end{align*}
The critical points are $x=-1,0,2$. Using, for instance, the test point method (which is the easiest method of solving an inequality), we obtain the following table.
$$\begin{array}{|c|c|c|c|c|c|c|c|} \hline x & x<-1 & -1 & -1<x<0 & 0 & 0<x<2 & 2 & x>2\\ \hline f’(x) & – & 0 & + & 0 & – & 0 & +\\ \hline f(x) & \searrow & f(-1) & \nearrow & f(0) & \searrow & f(2) &\nearrow\\ \hline \end{array}$$
So we find that $f$ is increasing on $(-1,0)\cup(2,\infty)$ and $f$ is decreasing on $(-\infty,-1)\cup(0,2)$.

Now, local maximum values and local minimum values can be identified by observing the change of sign of $f’(x)$ at each critical point.

Theorem. [The First Derivative Test] Suppose that $c$ is a critical point of a differentiable function $f(x)$.

1. If the sign of $f’(x)$ changes from $+$ to $-$ at $c$, $f(c)$ is a local maximum.
2. If the sign of $f’(x)$ changes from $-$ to $+$ at $c$, $f(c)$ is a local minimum.
3. If the sign $f’(x)$ does not change at $c$, $f$ has neither a local maximum nor a local minimum at $c$.

Example. In the previous example, the sign of $f’(x)$ changes from $+$ to $-$ at $0$, so $f(0)=5$ is a local maximum. The sign of $f’(x)$ changes from $-$ to $+$ at $-1$ and at $2$, so $f(-1)=0$ and $f(2)=-27$ are local minimum values.

The following figure confirms our findings from the above two examples.

The graph of f(x)=3x^4-4x^3-12x^2+5

The Second Derivative Test

The second order derivative $f^{\prime\prime}(x)$ can provide us an additional piece of information on $y=f(x)$, namely the concavity of the graph of $y=f(x)$.

Definition. If the graph of $f$ lies above all of its tangents on an open interval $I$, it is called concave upward on $I$. If the graph of $f$ lies below all of its tangents on $I$, it is called concave downward on $I$.

From here on, $\smile$ denotes “concave up” and $\frown$ denotes “concave down”.

Definition. A point $(d,f(d))$ on the graph of $y=f(x)$ is called a point of inflection if the concavity of the graph of $f$ changes from $\smile$ to $\frown$ or from $\frown$ to $\smile$ at $(d,f(d))$. The candidates for the points of inflection may be found by solving the equation $f^{\prime\prime}(x)=0$ as shown in the example below.

Theorem. [Concavity Test]

1. If $f^{\prime\prime}(x)>0$ for all x in an open interval $I$, the graph of $f$ is concave up on $I$.
2. If $f^{\prime\prime}(x)<0$ for all x in an open interval $I$, the graph of $f$ is concave down on $I$.

Theorem. [The Second Derivative Test] Suppose that $f’(c)=0$ i.e. $c$ is a critical point of $f$. Suppose that $f^{\prime\prime}$ is continuous near $c$.

1. If $f^{\prime\prime}(c)>0$ then $f(c)$ is a local minimum.
2. If $f^{\prime\prime}(c)<0$ then $f(c)$ is a local maximum.

Example. Let $f(x)=-x^4+2x^2+2$.

1. Find and identify all local maximum and local minimum values of $f(x)$ using the Second Derivative Test.
2. Find the intervals on which the graph of $f(x)$ is concave up or concave down. Find all points of inflection.

Solution. 1. First we find the critical points of $f(x)$ by solving the equation $f’(x)=0$:
$$f’(x)=-4x^3+4x=-4x(x^2-1)=-4x(x+1)(x-1)=0.$$ So $x=-1,0,1$ are critical points of $f(x)$ Next, $f^{\prime\prime}(x)=-12x^2+4$. Since $f^{\prime\prime}(0)=4>0$ and $f^{\prime\prime}(-1)=f^{\prime\prime}(1)=-8<0$, by the Second Derivative Test, $f(0)=2$ is a local minimum value and $f(-1)=f(1)=3$ is a local maximum value.

2. First we need to solve the equation $f”(x)=0$:
$$f^{\prime\prime}(x)=-12x^2+4=-12\left(x^2-\frac{1}{3}\right)=-12\left(x+\frac{1}{\sqrt{3}}\right)\left(x-\frac{1}{\sqrt{3}}\right)=0.$$ So $f^{\prime\prime}(x)=0$ at $x=\pm\displaystyle\frac{1}{\sqrt{3}}$. By using the test-point method we find the following table:
$$\begin{array}{|c||c|c|c|c|c|} \hline x & x<-\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{3}} & x>\frac{1}{\sqrt{3}}\\ \hline f^{\prime\prime}(x) & – & 0 & + & 0 & -\\ \hline f(x) & \frown & f\left(-\frac{1}{\sqrt{3}}\right)=\frac{23}{9} & \smile & f\left(\frac{1}{\sqrt{3}}\right)=\frac{23}{9} & \frown\\ \hline \end{array}$$
The graph of $f(x)$ is concave down on the intervals $\left(-\infty,-\frac{1}{\sqrt{3}}\right)\cup\left(\frac{1}{\sqrt{3}},\infty\right)$ and is concave up on the interval $\left(-\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}\right)$. The points of inflection are $\left(-\frac{1}{\sqrt{3}},\frac{23}{9}\right)$ and $\left(\frac{1}{\sqrt{3}},\frac{23}{9}\right)$.

The following figure confirms our findings from the above example.

The graph of f(x)=-x^4+2x^2+2 with points of inflection (in blue)

# The Substitution Rule

If given integration takes the form $\int f(g(x))g’(x)dx$ then it can be converted to a simpler integration that we may be able to evaluate by the substitution $u=g(x)$. In fact, the integration is given in terms of the new variable $u$ as
$$\int f(g(x))g’(x)dx=\int f(u)du.$$

Example. Evaluate $\int x\sqrt{1+x^2}dx$.

Solution. Let $u=1+x^2$. Then $du=2xdx$. So,
\begin{align*}
\int x\sqrt{1+x^2}dx&=\frac{1}{2}\int\sqrt{u}du\\
&=\frac{1}{3}u^{\frac{3}{2}}+C\\
&=\frac{1}{3}(1+x^2)^{\frac{3}{2}}+C,
\end{align*}
where $C$ is an arbitrary constant.

Example. Evaluate $\int\cos (7\theta+5)d\theta$.

Solution. Let $u=7\theta+5$. Then $du=7d\theta$. So,
\begin{align*}
\int\cos (7\theta+5)d\theta&=\frac{1}{7}\int\cos udu\\
&=\frac{1}{7}\sin u+C\\
&=\frac{1}{7}\sin(7\theta+5)+C,
\end{align*}
where $C$ is an arbitrary constant.

Example. Evaluate $\int x^2\sin(x^3)dx$.

Solution. Let $u=x^3$. Then $du=3x^2dx$. So,
\begin{align*}
\int x^2\sin(x^3)dx&=\frac{1}{3}\int \sin udu\\
&=-\frac{1}{3}\cos u+C\\
&=-\frac{1}{3}\cos(x^3)+C,
\end{align*}
where $C$ is an arbitrary constant.

How do we evaluate a definite integral of the form $\int_a^b f(g(x))g’(x)dx$? The following example shows you how.

Example. Evaluate $\int_{-1}^1 3x^2\sqrt{x^3+1}dx$.

Solution. First let us calculate the indefinite integral $\int 3x^2\sqrt{x^3+1}dx$. Let $u=x^3+1$. Then $du=3x^2dx$. So,
\begin{align*}
\int 3x^2\sqrt{x^3+1}dx&=\int\sqrt{u}du\\
&=\frac{2}{3}u^{\frac{3}{2}}+C\\
&=2(x^3+1)^{\frac{3}{2}}+C,
\end{align*}
where $C$ is an arbitrary constant. Now by Fundamental Theorem of Calculus,
\begin{align*}
\int_{-1}^1 3x^2\sqrt{x^3+1}dx&=\frac{2}{3}[(x^3+1)^{\frac{3}{2}}]_{-1}^1\\
&=\frac{4\sqrt{2}}{3}.
\end{align*}

But there is a better way to do this as shown in the following theorem. Its proof is straightforward.

Theorem. If $u’$ is continuous on $[a,b]$ and $f$ is continuous on the range of $u$, then
$$\int_a^bf(u(x))u’(x)dx=\int_{u(a)}^{u(b)}f(u)du.$$

Example. Let us replay the previous example using this theorem. Again let $u=x^3+1$. Then $du=3x^2dx$ and $u(-1)=0$, $u(1)=2$. Now, by the above theorem,
\begin{align*}
\int_{-1}^1 3x^2\sqrt{x^3+1}dx&=\int_0^2\sqrt{u}du\\
&=\frac{2}{3}[u^{\frac{3}{2}}]_0^2\\
&=\frac{4\sqrt{2}}{3}.
\end{align*}
I believe you will find this more simple than previous method.

I will finish this lecture with the following nice properties.

Theorem. Let $f$ be a continuous function on $[-a,a]$.

(a) If $f$ is an even function, then $\int_{-a}^a f(x)dx=2\int_0^a f(x)dx$.

(b) If $f$ is an odd function, then $\int_{-a}^a f(x)dx=0$.

This can be easily understood from pictures using the symmetries of even and odd functions. But the theorem can be proved using substitution. I will leave it to you.

# Mean Value Theorem

The following theorem is something one can easily picture intuitively.

Theorem. [Rolle's Theorem]
Let $f$ be continuous on the closed interval $[a,b]$ and differentiable on the open interval $(a,b)$. If $f(a)=f(b)$, then there exists a number $c$ in $(a,b)$ such that $f’(c)=0$.

Example. Show that the equation $x^3+x-1=0$ has exactly only one real root.

Solution. Let $f(x)=x^3+x-1$. Note that $f(0)=-1$ and $f(1)=1$. So by the Intermediate Value Theorem, we see that there exists at least a root of the equation $x^3+x-1=0$ in the interval $(0,1)$. Now suppose that there are two different roots $a$ and $b$ of the equation $x^3+x-1=0$. Without loss of generality, we may assume that $a<b$. Then $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. By Rolle’s Theorem then, there exist a number $c$ in $(a,b)$ such that $f’(c)=0$. However, $f’(x)=3x^2+1\geq 1$ for all real number $x$. This is a contradiction. Therefore, there should be only one root of the equation.

The graph of f(x)=x^3+x-1

Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Define $g(x)$ to be the distance between $f(x)$ and the line segment from $(a,f(a))$ to $(b,f(b))$, i.e.
$$g(x)=f(x)-\frac{f(b)-f(a)}{b-a}(x-a)-f(a).$$ Then $g(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Since $g(a)=g(b)=0$, by Rolle’s theorem there exists a number $c$ in $(a,b)$ such that $g’(c)=f’(c)-\frac{f(b)-f(a)}{b-a}=0$. Therefore, we proved the following theorem.

Mean Value Theorem

Theorem. [Mean Value Theorem]
Let $f$ be continuous on $[a,b]$ and differentiable on $(a,b)$. Then there exists a number $c$ in $(a,b)$ such that
$$f’(c)=\frac{f(b)-f(a)}{b-a}.$$

The following example is an application of the Mean Value Theorem.

Example. Suppose that $f(0)=-3$ and $f’(x)\leq 5$ for all values of $x$. How large can $f(2)$ possibly be?

Solution. By the Mean Value Theorem, there exists a number $c$ in $(0,2)$ such that
$$f’(c)=\frac{f(2)-f(0)}{2-0}=\frac{f(2)+3}{2}.$$
Since $f’(c)\leq 5$,
\begin{align*}
f(2)&=2f’(c)-3\\
&\leq 2\cdot 5-3=7.
\end{align*}
Hence, $7$ is the largest possible value of $f(2)$.

Using Mean Value Theorem, one can prove the following theorem.

Theorem. If $f’(x)=0$ for all $x$ in the open interval $(a,b)$, then $f$ is constant on $(a,b)$.

# Maximum and Minimum

Maximum and Minimum

There are two different types of extremum (maximum or minimum) values of a function $y=f(x)$. We may consider a value of $y$ that is an extremum globally on the domain or we may also consider a value of $y$ that is an extremum locally around an $x$ value.

A function $f$ has an absolute maximum at $c$ if $f(c)\geq f(x)$ for all $x$ in the domain of $f$. Similarly, $f$ has an absolute minimum at $c$ if $f(c)\leq f(x)$ for all $x$ in the domain of $f$.

A function $f$ has a local maximum (or relative maximum) at $c$ if $f(c)\geq f(x)$ in some neighborhood of $c$ (i.e an open interval that contains $c$). Similarly, $f$ has a local minimum (or relative minimum) at $c$ if $f(c)\leq f(x)$ in some neighborhood of $c$.

Example.

The graph of f(x)=3x^4-16x^3+18x^2 on [-1,4]

The above figure shows the graph of $f(x)=3x^4-16x^3+18x^2$, $-1\leq x\leq 4$. It has a local maximum at $x=1$ and a local minimum at $x=3$. The local minimum $f(3)=-27$ is also an absolute minimum. $f$ has an absolute maximum $f(-1)=37$. This $f(-1)=37$ is not a local maximum by the way. The reason is that there is no local neighborhood around $x=-1$ as the domain is given by $[-1,4]$.

A natural question one may ask is whether a function always has an absolute maximum and an absolute minimum. You can easily find many examples that show that a function does not necessarily have an absolute maximum or an absolute minimum value. For instance, $y=x$ on $(-\infty,\infty)$ has neither an absolute maximum nor an absolute minimum. The function $y=x^2$ on $[0,1)$ has an absolute minimum 0 at $x=0$ but has no absolute maximum.

Theorem. [Max-Min Theorem, Fermat]
If $f$ is continuous on a closed interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$.

The following theorem is also due to Fermat.

Theorem. If $f$ has a local maximum or a local minimum at $c$ and if $f’(c)$ exists, then $f’(c)=0$.

The converse of this theorem is not necessarily true i.e. $f’(c)=0$ does not necessarily mean that $f(c)$ is a local maximum or a local minimum. For example, consider $f(x)=x^3$. $f’(0)=0$ but $f(x)$ has neither a local maximum nor a local minimum at $x=0$ as shown in figure below.

The graph of f(x)=x^3

The above theorem is important as an absolute maximum and an absolute minimum may be found among local maximum values, local minimum values and the evaluations of $f$ at the end points, $f(a)$ and $f(b)$. To find local maximum values and local minimum values, we first find points $c$ such that $f’(c)=0$. Such points are called critical points. The reason they are called critical points is that the graph of a function changes from increasing to decreasing or from decreasing to increasing at a critical point.

Definition. A critical point of a function $f(x)$ is a number $c$ in the domain of $f$ such that either $f’(c)=0$ or $f’(c)$ does not exist.

Recipe of Finding Absolute Maximum and Absolute Minimum

Let $f$ be a continuous function on a closed interval $[a,b]$.

Step 1. Find all critical points of $f$ in $(a,b)$.

Step 2. Evaluate $f$ at each critical point obtained in Step 1.

Step 3. Find $f(a)$ and $f(b)$.

Step 4. Compare all the values obtained in Steps 2 and 3. The largest value is the absolute maximum and the smallest value is the absolute minimum.

Example. Find the absolute maximum and the absolute minimum values of
$$f(x)=x^3-3x^2+1,\ -\frac{1}{2}\leq x\leq 4.$$

Solution.

Step 1. Find all critical points of $f$ in $\left(-\frac{1}{2},4\right)$.

$f’(x)=3x^2-6x$. Set $f’(x)=0$ i.e. $3x^2-6x=0$. $3x^2-6x$ is factored as $3x(x-2)$. So we find two critical points $0, 2$.

Step 2. Evaluate $f$ at each critical point obtained in Step 1.

$f(0)=1$ and $f(2)=-3$.

Step 3. Find $f\left(-\frac{1}{2}\right)$ and $f(4)$.

$f\left(-\frac{1}{2}\right)=\frac{1}{8}$ and $f(4)=17$.

Step 4. Compare all the values obtained in Steps 2 and 3.

The largest value is $f(4)=17$ so this is the absolute maximum value of $f$ on $\left[-\frac{1}{2},4\right]$. The smallest value is $f(2)=-3$ so this is the absolute minimum of $f$ on $\left[-\frac{1}{2},4\right]$.

# Linear Approximations and Differentials

Linear Approximation

Let $y=f(x)$ be a differentiable function. The function $f(x)$ can be approximated by the tangent line to $y=f(x)$ at $a$ if $x$ is near $a$. Such an approximation is called a linear approximation.

If $x\approx a$ then $\Delta x=x-a\approx 0$, so we have
\begin{align*}
\frac{\Delta y}{\Delta x}&\approx \frac{dy}{dx}\\
&=f’(a).
\end{align*}
This means that
$$\frac{f(x)-f(a)}{x-a}\approx f’(a),$$
i.e.

\label{eq:lineapprox}
f(x)\approx f(a)+f’(a)(x-a).

The equation \eqref{eq:lineapprox} is called the linear approximation or tangent line approximation of $f$ at $a$. The linear function

L(x):=f(a)+f’(a)(x-a)

is called the linearization of $f$ at $a$. Notice that $L(x)$ is the equation of tangent line to $f$ at $a$.

Example. Find the linearlization of $f(x)=\sqrt{x+3}$ at $a=1$ and use it to approximate $\sqrt{3.98}$ and $\sqrt{4.05}$.

Solution.

Linear approximation of f(x)=sqrt(x+3) at a=1

$f’(x)=\frac{1}{2\sqrt{x+3}}$, so
\begin{align*}
L(x)&=f(1)+f’(1)(x-1)\\
&=2+\frac{1}{4}(x-1)\\
&=\frac{x}{4}+\frac{7}{4}.
\end{align*}
When $x\approx 1$, we have the approximation
$$\sqrt{x+3}\approx \frac{x}{4}+\frac{7}{4}.$$
$\sqrt{3.98}$ can be written as $\sqrt{3+0.98}$. Hence,
\begin{align*}
\sqrt{3.98}&\approx \frac{0.98}{4}+\frac{7}{4}\\
&=1.995.
\end{align*}
$\sqrt{4.05}$ can be written as $\sqrt{4+1.05}$. Hence,
\begin{align*}
\sqrt{4.05}&\approx \frac{1.05}{4}+\frac{7}{4}\\
&=2.0125.
\end{align*}

Differentials

Differentials

As seen in the above figure, when $\Delta x\approx 0$, $\Delta x=dx$ and $\Delta y\approx dy$. On the other hand, $\frac{dy}{dx}=f’(x)$. Hence, we obtain

\label{eq:differential}
\Delta y\approx f’(x)\Delta x.

Example. The radius of a sphere was measured and found to be 21 cm with a possible error in measurement of at most 0.05 cm. What is the maximum error in using this value of the radius to compute the volume of the sphere?

Solution. Let $V$ denote the volume of a sphere of radius $r$. Then $V=\frac{4}{3}\pi r^3$. What we are trying to find is $\Delta V$ with $\Delta r=0.05$ cm. As seen in \eqref{eq:differential}, $\Delta V\approx dV$, so we find $dV$ instead because finding $dV$ is easier than findingthe exact error $\Delta V$. Differentiating $V$ with respect to $r$, we obtain
\begin{align*}
dV&=4\pi r^2 dr\\
&=4\pi r^2\Delta r\\
&=4\pi\cdot(21)^2\cdot 0.05\\
&=277.
\end{align*}
So the maximum error in the calculated volume is about 277 $\mbox{cm}^3$.

# Implicit Differentiation

A lot of time we have seen functions defined as $y=f(x)$. This clearly shows that $y$ is a function of the independent variable $x$. But often functions are defined implicitly. For instance, consider the equation $x^2+y^2=25$. Of course this is the equation of circle centered at the center $(0,0)$ with radius $5$. Also circles are not functions. But if we say $y\geq 0$, then the equation describes the upper half-circle which is a function defined by $y=\sqrt{25-x^2}$. Functions defined by equations like $x^2+y^2=25$ are called implicit functions. In some cases like $x^2+y^2=25$, we can easily write an implicit function explicitly as $y=f(x)$, but in many cases we cannot. For example, $x^3+y^3=6xy$. So, we need to devise a way to differentiate an implicit function without writing it as $y=f(x)$. This can indeed be done by the chain rule. You just assume that $y$ is a function of $x$ and use the chain rule. For example,
\begin{align*}
\frac{d}{dx}y^n&=(y^n)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\
&=ny^{n-1}\frac{dy}{dx}.
\end{align*}
Let us take a look at another example.
\begin{align*}
\frac{d}{dx}\cos y&=(\cos y)’\frac{dy}{dx}\ (y\ \mbox{is the innermost function})\\
&=-\sin y\frac{dy}{dx}.
\end{align*}
Here come more examples.

Example. If $x^2+y^2=25$, find $\frac{dy}{dx}$.

Solution. Differentiating the equation with respect to $x$, we obtain
$$2x+2y\frac{dy}{dx}=0.$$
Solving the resulting equation for $\frac{dy}{dx}$, we obtain
$$\frac{dy}{dx}=-\frac{x}{y}.$$

Example.

1. Find $y’$ if $x^3+y^3=6xy$.

Solution. Differentiate the equation with respect to $x$. Then we obtain
$$3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}.$$
Solving the resulting equation for $\frac{dy}{dx}$, we obtain
$$\frac{dy}{dx}=\frac{2y-x^2}{y^2-2x}.$$

2. Find the tangent to $x^3+y^3=6xy$ at $(3,3)$.

Solution. The equation of tangent is
$$y-3=\left[\frac{dy}{dx}\right]_{(3,3)}(x-3).$$
$$\left[\frac{dy}{dx}\right]_{(3,3)}=\frac{2\cdot 3-(3)^2}{3^2-2\cdot 3}=-1.$$ Therefore, the tangent is given by $y=-x+6$.

# The Chain Rule

Let us consider the function $y=\sqrt{x^2+1}$. Notice that this is a composite function $y=\sqrt{u}$ and $u=x^2+1$. In general, a composite function can be written as $y=f(u)$ where $u$ is a function of $x$, $u=g(x)$. While we know how to differentiate $y=\sqrt{u}$ (i.e. finding $\frac{dy}{du}$) and $u=x^2+1$ (i.e. finding $\frac{du}{dx}$), we do not know how to differentiate $y=\sqrt{x^2+1}$ (i.e finding $\frac{dy}{dx}$). In this lecture, we would like to devise a way to differentiate a composite function. This is actually very important because the differentiable functions we stumble onto most of time are composite functions.

Let $y=f(u)$ and $u=g(x)$ and assume that both $\frac{dy}{du}$ and $\frac{du}{dx}$ exist. Now,
\begin{align*}
\frac{\Delta y}{\Delta x}&=\frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x}\\
&=\frac{f(u+\Delta u)-f(u)}{\Delta u}\cdot\frac{g(\Delta x+x)-g(x)}{\Delta x}.
\end{align*}
Hence,
\begin{align*}
\frac{dy}{dx}&=\lim_{\Delta x\to 0}\frac{\Delta y}{\Delta x}\\
&=\lim_{\Delta u\to 0}\frac{\Delta y}{\Delta u}\cdot\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x}\ (\Delta u\to 0\ \mbox{as}\ \Delta x\to 0)\\
&=\frac{dy}{du}\cdot\frac{du}{dx}
\end{align*}
or
\begin{align*}
\frac{dy}{dx}&=\lim_{\Delta u\to 0}\frac{f(u+\Delta u)-f(u)}{\Delta u}\cdot\lim_{\Delta x\to 0}\frac{g(\Delta x+x)-g(x)}{\Delta x}\\
&=f’(u)g’(x).
\end{align*}

Theorem. [The Chain Rule]
Let $y=f(u)$ and $u=g(x)$. If both $\frac{dy}{du}$ and $\frac{du}{dx}$ exist, then $\frac{dy}{dx}$ exists and
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=f’(u)g’(x).
\end{align*}

Remark. The derivation of the chain rule shown above is not rigorously correct. The reason is that $\Delta u$ may become $0$. There is a more rigorous proof of the chain rule but we will not discuss that here.

Remark. Students commonly feel a difficulty with applying the chain rule when they learn it for the first time. The difficulty usually is not about understanding the chain rule itself but identifying the function $u=g(x)$. The candidate for $u$ is usually the function inside parentheses (or brackets) or the innermost function.

Example. We are now ready to find $\frac{dy}{dx}$ when $y=\sqrt{x^2+1}$. In this case, we don’t see parentheses or brackets but the innermost function is $x^2+1$. Let $u=x^2+1$. Then $y=\sqrt{u}$. Now,
\begin{align*}
\frac{dy}{du}&=\frac{1}{2\sqrt{u}}\\
&=\frac{1}{2\sqrt{x^2+1}},\\
\frac{du}{dx}&=2x.
\end{align*}
so, we have by the chain rule
$$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=\frac{x}{\sqrt{x^2+1}}.$$

Example. Differentiate $y=(x^3-1)^{100}$.

Solution. The function inside parentheses is $x^3-1$. So, it is our candidate. Let $u=x^3-1$. Then $y=u^{100}.$
By the chain rule,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=100u^{99}\cdot(3x^2)\\
&=300x^2(x^3-1)^{99}.
\end{align*}

Example. Find the derivative of each function.

1. $y=\sin 4x$.

Solution. The innermost function is $4x$. Let $u=4x$. Then $y=\sin u$. By the chain rule,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=\cos u\cdot4\\
&=4\cos 4x.
\end{align*}

2. $y=\sqrt{\sin x}$.

Solution. The innermost function is $\sin x$. Let $u=\sin x$. Then $y=\sqrt{u}$. By the chain rule,
\begin{align*}
\frac{dy}{dx}&=\frac{dy}{du}\cdot\frac{du}{dx}\\
&=\frac{1}{2\sqrt{u}}\cdot\cos x\\
&=\frac{\cos x}{2\sqrt{\sin x}}.
\end{align*}

# The Fundamental Theorem of Calculus

First we begin with the Mean Value Theorem for Definite Integrals.

Theorem. If $f$ is continuous on $[a,b]$, then at some point $c\in [a,b]$,
$$f(c)(b-a)=\int_a^b f(x)dx$$
or

\label{eq:mvt}
f(c)=\frac{1}{b-a}\int_a^b f(x)dx.

Notice that the RHS of \eqref{eq:mvt} is the average of $f(x)$ on $[a,b]$.

Example. Find the average value of $f(x)=4-x$ on $[0,3]$ and $c\in[0,3]$ at which $f$ actually takes on this value.

Solution. The region under $y=4-x$ on $[0,3]$ is a trapezoid and its area is $\frac{15}{2}$. So,
\begin{align*}
av(f)&=\frac{1}{3-0}\int_0^3(4-x)dx\\
&=\frac{5}{2}.
\end{align*}
Let $f(c)=\frac{5}{2}$. Then $4-c=\frac{5}{2}$ and $c=\frac{3}{2}$.

Example. Show that if $f$ is continuous on $[a,b]$ ($a\ne b$) and if $\int_a^b f(x)=0$ then $f(x)=0$ at least once in $[a,b]$.

Solution. By the MVT, there exists $c\in[a,b]$ such that
$$f(c)=\frac{1}{b-a}\int_a^b f(x)dx=0.$$

Suppose that $f(t)$ is an integrable function on a finite intervale $I$. Let $a\in I$. Then
$$F(x):=\int_a^x f(t)dt$$
defines a fuction on the interval $I$. Let $f(x)$ be a continuous function on $[a,b]$. Then

Claim 1: $F(x)=\int_a^x f(t)dt$ is continuous on $[a,b]$. (For a proof click here. For those who are not familiar with $\epsilon$-$\delta$ argument, an alternative proof is available here.)

Claim 2: $F(x)$ is differentiable on $(a,b)$ and $F’(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$. (For a proof click here.)

The claims 1 and 2 constitute:

The Fundamental Theorem of Calculus (Part I)

If $f$ is continuous on $[a,b]$, then $F(x)=\int_a^xf(t)dt$ is continuous on $[a,b]$ and is differentiable on $(a,b)$. Moreover, $F’(x)=\frac{d}{dx}\int_a^xf(t)dt=f(x)$.

The Fundamental Theorem of Calculus (Part II) relates the definitel integral and an antiderivative of a function $f(x)$.

If $f$ is continuous on $[a,b]$ and $F$ is any antiderivative of $f$ on $[a,b]$, then
$$\int_a^b f(x)dx=F(b)-F(a).$$(For a proof click here.)

Note: The usual notation for $F(b)-F(a)$ is $F(x)|_a^b$ or $[F(x)]_a^b$.