Unit 4: Techniques of Integration
Preface
This is the note of the MIT Course 18.01.2 Calculus. For more details, refer to MIT 18.x Catalog, and it can also be found at MIT Open Learning Library.
Integrals of Trig Powers
Trigonometric Identities
The Pythagorean Theorem
$$ \begin{aligned} \cos^2(\theta)+\sin^2(\theta)\quad=&\quad1\\ 1+\tan^2(\theta)\quad=&\quad\sec^2(\theta)\qquad\left(\theta\ne\frac{\pi}2+n\pi\right)\\ \cot^2(\theta)+1\quad=&\quad\csc^2(\theta)\qquad(\theta\ne n\pi) \end{aligned} $$
The Angle Sum Formulas
$$ \begin{aligned} \sin(a+b)\quad=&\quad\sin(a)\cos(b)+\sin(b)\cos(a)\\ \cos(a+b)\quad=&\quad\cos(a)\cos(b)-\sin(a)\sin(b) \end{aligned} $$
The Half Angle Formulas
$$ \begin{aligned} \sin^2(a)\quad=&\quad\frac{1-\cos(2a)}2\\ \cos^2(a)\quad=&\quad\frac{1+\cos(2a)}2 \end{aligned} $$
Integrals of Products of Powers of Sin and Cosine
One Exponent Equals One
$$ \begin{aligned} \int\cos^m(x)\sin(x)\ dx\quad=&\begin{cases} \displaystyle-\frac{\cos^{m+1}(x)}{m+1}+C,\quad\text{if $m\ne -1$}\\ -\ln\vert\cos(x)\vert+C,\quad\text{if $m=-1$} \end{cases}\\ \int\sin^m(x)\cos(x)\ dx\quad=&\begin{cases} \displaystyle\frac{\sin^{m+1}(x)}{m+1}+C,\quad\text{if $m\ne -1$}\\ \ln\vert\sin(x)\vert+C,\quad\text{if $m=-1$} \end{cases} \end{aligned} $$
When $m=-1$, $$ \begin{aligned} \int\tan(x)\ dx\quad=&\quad-\ln\vert\cos(x)\vert+C\\ \int\cot(x)\ dx\quad=&\quad\ln\vert\sin(x)\vert+C \end{aligned} $$ When $m=-2$, $$ \begin{aligned} \int\tan(x)\sec(x)\ dx\quad=&\quad\sec(x)+C\\ \int\cot(x)\csc(x)\ dx\quad=&\quad-\csc(x)+C \end{aligned} $$
One Postive Odd Exponent
Let $2m+1$ be any positive integer and $n$ any real number. Then we can use substitution: $$ \begin{aligned} \int\cos^{2m+1}(x)\sin^n(x)\ dx\quad:&\quad\text{use }u=\sin(x)\\ \int\sin^{2m+1}(x)\cos^n(x)\ dx\quad:&\quad\text{use }u=\cos(x) \end{aligned} $$
Two Positive Even Exponents
To evaluate $$ \int\sin^{2m}(x)\cos^{2n}(x)\ dx\qquad 2m,2n\ge0\text{ and are even} $$ apply the half angle formulas to the integrand as many times as needed.
Integrals of Products of Powers of Tangent and Cosecant
Odd Positive Exponent of Tangent
The integrals of the form $$ \int\tan^{2m+1}(x)\sec^n(x)\ dx\qquad\text{$n$ real, $2m+1\ge1$ odd positive integer} $$
$$ \int\cot^{2m+1}(x)\csc^n(x)\ dx\qquad\text{$n$ real, $2m+1\ge1$ odd positive integer} $$
can be evaluated using the substitution $u=\sec(x)$, and $u=\csc(x)$ respectively.
Alternatively, these can also be written as $$ \int\sin^{2m+1}(x)\cos^{-n}(x)\ dx\qquad\text{$n$ real, $2m+1\ge1$ odd positive integer} $$
$$ \int\cos^{2m+1}(x)\sin^{-n}(x)\ dx\qquad\text{$n$ real, $2m+1\ge1$ odd positive integer} $$
we can also evaluate them using the substitution $u=\cos(x)$ and $u=\sin(x)$ respectively.
Even Positive Exponent of Secant
Let $2m\ge2$ be a positive even integer, and $n$ any real number.
We can use $u=\tan(x)$ to evaluate $$ \int\sec^{2m}(x)\tan^n(x)\ dx\qquad\text{$2m\ge2$ even integer, $n$ real} $$ and use $u=\cot(x)$ to evaluate $$ \int\csc^{2m}(x)\cot^n(x)\ dx\qquad \text{$2m\ge2$ even integer, $n$ real} $$ These are equivalent to $$ \int\sin^p(x)\cos^q(x)\ dx\qquad\text{for $p$, $q$ real, $p+q=-2k$ even negative integer} $$
Integrals Table
$$ \begin{aligned} \int\sec^2(x)\ dx\quad=&\quad\tan(x)+C\\ \int\csc^2(x)\ dx\quad=&\quad-\cot(x)+C\\ \int\sec(x)\tan(x)\ dx\quad=&\quad\sec(x)+C\\ \int\csc(x)\cot(x)\ dx\quad=&\quad-\csc(x)+C\\ \int\tan(x)\ dx\quad=&\quad-\ln\vert\cos(x)\vert+C\\ \int\cot(x)\ dx\quad=&\quad\ln\vert\sin(x)\vert+C\\ \int\sec(x)\ dx\quad=&\quad\ln\vert\sec(x)+\tan(x)\vert+C\\ \int\csc(x)\ dx\quad=&\quad-\ln\vert\csc(x)+\cot(x)\vert+C \end{aligned} $$
Trig Integrals and Binomial Expansion
Binomial Coefficients
The binomial theorem says that
$$
(1+y)^m=\sum_{k=0}^m\dbinom{m}{k}y^k=\sum_{k=0}^m\frac{m!}{(m-k)!\ k!}y^k
$$
where $m!$ is called $m$ factorial.
And we get the coefficient of the $y^k$ to be
$$
\dbinom{m}{k}=\frac{m!}{(m-k)!\ k!}
$$
which is called the binomial coeffecient, and $\dbinom{m}{k}$ is read as m choose k.
General Formula for Integral
Integral with One Positive Odd Exponent of Sine or Cosine
Let $2m+1$ be any positive integer and $n$ any real number such that $n\ne-(2k+1)$ for any integer $k$ such that $0\le k\le m$, we obtain the following formulas: $$ \begin{aligned} \int\cos^{2m+1}(x)\sin^n(x)\ dx\ =&\ \sum_{k=0}^m(-1)^k\dbinom{m}{k}\frac{\sin^{2k+n+1}(x)}{2k+n+1}+C\\ \int\sin^{2m+1}(x)\cos^n(x)\ dx\ =&\ \sum_{k=0}^m(-1)^{k+1}\dbinom{m}{k}\frac{\cos^{2k+n+1}(x)}{2k+n+1}+C \end{aligned} $$
Integral with Even Positive Power of Secant
$$ \begin{aligned} \sec^{2m}(x)\ =&\ \sec^2(x)(1+\tan^2(x))^{m-1}\\ =&\ \sec^2(x)\left(\sum_{k=0}^{m-1}\dbinom{m-1}k\tan^{2k}(x)\right)\\ \int\sec^{2m}(x)\ dx\ =&\ \int\left(\sum_{k=0}^{m-1}\dbinom{m-1}ku^{2k}\right)\ du\qquad\text{use $u=\tan(x)$}\\ =&\ \sum_{k=0}^{m-1}\dbinom{m-1}k\frac{\tan^{2k+1}(x)}{2k+1}+C \end{aligned} $$
Trig Substitution
An Example
- Consider the integral:
$$ \int\frac{dx}{x^2\sqrt{1+x^2}} $$
- Identify this part:
$$ \sqrt{1+x^2} $$
- Make a substitution to simplify this part based on $\sec^2\theta=1+\tan^2\theta$.
$$ \begin{cases} x=\tan\theta\\ \sqrt{1+x^2}=\sqrt{\sec^2\theta}=\sec\theta\\ dx=\sec^2\theta\ d\theta \end{cases} $$
- Plug into integral:
$$ \int\frac{dx}{x^2\sqrt{1+x^2}}=\int\frac{\sec\theta\ d\theta}{\tan^2\theta} $$
- Rewrite in terms of $\sin$ and $\cos$:
$$ =\int\frac{\cos\theta\ d\theta}{\sin^2\theta} $$
- Try substitution $\displaystyle\begin{cases}u=\sin\theta\\ du=\cos\theta\ d\theta\end{cases}$
$$ =\int\frac{du}{u^2} $$
- Evaluate it,
$$ =\frac{-1}u+C $$
- Then back substitution by replace $u$ by $\sin\theta$
$$ =\frac{-1}{\sin\theta}+C $$
- As $\theta=\arctan x$, we can evaluate $\sin\theta$ by drawing diagrams.
Therefore,
$$ \frac{1}{\sin\theta}=\frac{\sqrt{1+x^2}}x $$
- Rewrite in terms of $x$:
$$ =-\frac{\sqrt{1+x^2}}{x}+C $$
Summary of Trig Substitutions
| If integrand contains: | Make Substituion: | To get: |
|---|---|---|
| $\sqrt{a^2-x^2}$ | $x=\begin{cases}a\cos\theta\\ a\sin\theta\end{cases}$ | $\sqrt{a^2-x^2}=\begin{cases}a\sin\theta\\ a\cos\theta\end{cases}$ |
| $\sqrt{a^2+x^2}$ | $x=a\tan\theta$ | $\sqrt{a^2+x^2}=a\sec\theta$ |
| $\sqrt{x^2-a^2}$ | $x=a\sec\theta$ | $\sqrt{x^2-a^2}=a\tan\theta$ |
Partial Fractions
Rational Functions
Ploynomials are function $P(x)=a_nx^n+a{n-1}x^{n-1}+\cdots+a_1x+a_0$ where the constants $a_i$ are real numbers and the powers $n$ are non-negative, finite integers.
The degree of a polynomial $P(x)$ is the integer $\deg(P)=n$ of the higest order $x$ term in the polynomial $P(x)$.
Rational functions are ratios of polynomials.
Method of Partial Functions
Given a rational function $\displaystyle\frac{P(x)}{Q(x)}$
For instance, we have a ration function
$$ \frac{4x-1}{x^2+x-2} $$
- Factor the denomiator $Q(x)$
$$ Q(x)=x^2+x-2=(x-1)(x+2) $$
- Set-up
$$ \frac{4x-1}{(x-1)(x+2)}=\frac A{x-1}+\frac B{x+2} $$
-
Solve for $A$ and $B$
- Multiply by $(x-1)$ and take the limit as $x\to 1$ $$ A=\frac{4-1}{1+2}=1 $$
- Multiply by $(x+2)$ and take the limit as $x\to-2$ $$ B=\frac{-8-1}{-2-1}=3 $$
-
Rewrite rational function
$$ \frac{4x-1}{x^2+x-2}=\frac 1{x-1}+\frac 3{x+2} $$
Cover-up Method
For a ration function $\frac{P(x)}{Q(x)}$ where $\deg(P)<\deg(Q)$, then the following is the fastest way:
- Factor $Q(x)$
- Set-up
- Cover-up
Repeated Linear Factors
For a ration function $\frac{P(x)}{Q(x)}$ where $\deg(P)<\deg(Q)$. If $Q(x)=(x-a)^n$, then the set-up for partial fractions is as follows:
$$ \frac{P(x)}{Q(x)}=\frac{A_1}{x-a}+\frac{A_2}{(x-a)^2}+\cdots+\frac{A_n}{(x-a)^n} $$
Quadratic Factors in the Demoniator
For a ration function $\frac{P(x)}{Q(x)}$ where $\deg(P)<\deg(Q)$. If $Q(x)=x^2+1$, then the set-up for partial fractions is as follows:
$$ \frac{P(x)}{Q(x)}=\frac{Bx+C}{x^2+1} $$
Improper Fractions
For a rational function $\frac{P(x)}{Q(x)}$ such that $\deg(P)\ge\deg(Q)$:
- Use Long Division to divide $Q(x)$ into $P(x)$ to obtain
$$ \frac{P(x)}{Q(x)}=p(x)+\frac{r(x)}{Q(x)} $$
where $p(x)$ and $r(x)$ are polynomials, and $\deg(r)<\deg(Q)$.
- Use the method of partial fractions to divide $\frac{r(x)}{Q(x)}$ into easier pieces.
Integration by Parts
We have already known the product rule of differentiation, which is:
$$ (u\ v)^{\prime}\quad=\quad u\ v^{\prime}+v^{\prime}\ u $$
Integration by parts comes from the product rule of differentiation.
$$ \int u\ v^{\prime}\ dx\quad=\quad u\ v-\int u^{\prime}\ v\ dx $$
Example 1:
$$ \int\ln(x)\ dx=\int\ln(x)\cdot 1\ dx $$
Let $u=\ln(x)$ and $v^{\prime}=1$, then $\displaystyle u^{\prime}=\frac 1 x$ and $v=x$,
then,
$$ \begin{aligned} =&\int u\ v^{\prime}\ dx\\ =&x\ln(x)-\int \frac 1 x\cdot x\ dx\\ =&x\ln(x)-x+C \end{aligned} $$
We can have more general formula:
$$ \begin{aligned} &\int x^p\ln(x)\ dx \qquad (p\ne0)\\ =& \int u^{\prime}\ v\ dx\qquad (u^{\prime}=x^p,\ v=\ln(x))\\ =&\ u\ v-\int u\ v^{\prime}\ dx\qquad (u=\frac{x^{p+1}}{p+1},\ v^{\prime}=\frac 1 x)\\ =&\ \frac{x^{p+1}}{p+1} \ln(x)-\frac{x^{p+1}}{(p+1)^2}+C \end{aligned} $$
Example 3:
Evaluate $\displaystyle \int x^3\cos(x^2)\ dx$
Let $$ u=x^2,\quad v^{\prime}=x\cos(x^2) $$
then we have $$ u^{\prime}=2x,\quad v=\frac{\sin(x^2)}2 $$
Therefore, $$ \begin{aligned} \int x^3\cos(x^2)\ dx=&\int u\ v^{\prime}\ dx\\ =&\ u\ v-\int u^{\prime}\ v\ dx\\ =&\ \frac{x^2\sin(x^2)}2+\frac{cos(x^2)}2+C \end{aligned} $$
Reduction Formula
Example 4:
We can find the reduction formula for $\displaystyle\int x^p(\ln x)^n\ dx$, where $n\ge 0$ integer, and $p\ne -1$ integer.
Define $$ I_n\ =\ \int x^p (\ln x)^n\ dx $$
Then we have the relationship between $I_n$ and $I_{n-1}$: $$ I_n\ =\ \frac{(\ln x)^n\ x^{p+1}}{p+1}-\frac{n}{p+1}\ I_{n-1} $$ Ignore the constant $C$ because we can add it at the end.
Example 5:
We will see the original integral reappears on the right hand side of the formula after applying integral by parts.
Evaluate $\displaystyle\int e^x\cos(x)\ dx$
Let $u=e^x$ and $v^{\prime}=\cos(x)$, then $u^{\prime}=e^x$ and $v=\sin(x)$.
We have $$ \begin{aligned} \int e^x\cos(x)\ dx\ =&\ \int u\ v^{\prime}\ dx\\ =&\ u\ v-\int u^{\prime}\ v\ dx\\ =&\ e^x\sin(x)-\int e^x\sin(x)\ dx \end{aligned} $$
Now we have to evaluate $\displaystyle\int e^x\sin(x)\ dx$.
Let $m=e^x$ and $n^{\prime}=\sin(x)$, then $m^{\prime}=e^x$ and $n=-\cos(x)$.
We have $$ \begin{aligned} \int e^x\sin(x)\ dx=&\ m\ n-\int m^{\prime}\ n\ dx\\ =&\ -e^x\cos(x)+\int e^x\cos(x)\ dx \end{aligned} $$
Therefore,
$$ \begin{aligned} \int e^x\cos(x)\ dx+\int e^x\sin(x)\ dx\ =&\ e^x\sin(x)-\int e^x\sin(x)\ dx+e^x\cos(x)+\int e^x\cos(x)\ dx\\ 2\int e^x\sin(x)\ dx\ =&\ e^x\sin(x)+e^x\cos(x)\\ \int e^x\sin(x)\ dx\ =&\ \frac{e^x\sin(x)+e^x\cos(x)}2 \end{aligned} $$
Example 6:
Now we will find the reduction formula of Cosine.
Define $$ C_n\ =\ \int\cos^n(x)\ dx $$
Using $\sin^2(x)+\cos^2(x)=1$, we get $$ C_n\ =\ C_{n-2}-\int \cos^{n-2}(x)\sin^2(x)\ dx $$
Use integration by parts where $n\ne 1$: $$ =\ \frac{\sin(x)\cos^{n-1}(x)}n+\frac{n-1}n\ C_{n-2} $$
Example 7:
We can find a reduction formula for the integral of odd positive powers of secant, but now we derive it from the reduction formula for the integral of powers of cosine.
We have already defined $C_n$, and now we define $$ \zeta_m\ =\ \int\sec^m(x)\ dx $$
When $m=n$, notice that $$ \zeta_m\ =\ C_{-n} $$
In example 6, we have $$ C_n\ =\ \frac{\sin(x)\cos^{n-1}(x)}n+\frac{n-1}n\ C_{n-2} $$
Replace $n$ by $-n$, we get $$ C_{-n}\ =\ \frac{\sin(x)\cos^{-n-1}(x)}{-n}+\frac{-n-1}{-n}\ C_{-n-2} $$
then $$ \begin{aligned} \zeta_m\ =&\ \frac{\sin(x)\cos^{-m-1}(x)}{-m}+\frac{-m-1}{-m}\ \zeta_{m+2}\\ =&\ \frac{m+1}{m}\ \zeta_{m+2}-\frac{\tan(x)\sec^m(x)}{m} \end{aligned} $$
Move $\zeta_{m+2}$ to the left hand side, $$ \begin{aligned} \zeta_{m+2}\ =&\ \frac{m}{m+1}\left(\zeta_m+\frac{\tan(x)\sec^m(x)}{m}\right)\\ =&\ \frac{m}{m+1}\ \zeta_m+\frac{\tan(x)\sec^m(x)}{m+1} \end{aligned} $$
Therefore, $$ \zeta_{m}\ =\ \frac{m-2}{m-1}\ \zeta_{m-2}+\frac{\tan(x)\sec^{m-2}(x)}{m-1} $$
Arc Length and Surface Area
Arc Length
Consider a small piece of a curve $\Delta s$.
We can use the length of secant line to approximate the length of this small curve: $$ \Delta s\ \approx\ \sqrt{\left(\Delta x\right)^2+\left(\Delta y\right)^2} $$
In differential, we get $$ ds\ =\ \sqrt{\left(dx\right)^2+\left(dy\right)^2} $$
For a curve $y=f(x)$, we can write $$ \begin{aligned} ds\ =&\ \sqrt{dx^2+dy^2}\\ =&\ \sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx\ =\ \sqrt{1+\big(f^{\prime}(x)\big)^2}\ dx\\ =&\ \sqrt{\left(\frac{dx}{dy}\right)^2+1}\ dy\ =\ \sqrt{\Big(\big(f^{-1}\big)^{\prime}(y)\Big)^2+1}\ dy \end{aligned} $$
Now, we can write the arc length $s$ of $y=f(x)$ over the interval shown by the integral:
$$
\begin{aligned}
\text{Arc Length }s\ =&\ \int_{s_0}^{s_n}\ ds\\
=&\ \int_{x_0}^{x_n}\sqrt{1+\big(f^{\prime}(x)\big)^2}\ dx\\
=&\ \int_{y_0}^{y_n}\sqrt{1+\Big(\big(f^{-1}\big)^{\prime}(y)\Big)^2}\ dy
\end{aligned}
$$
Surface Area
Consider the surface area created by rotating the curve $y=f(x)$ about the $x$-axis.
We can compute the surface area $A$ by the integral: $$ \begin{aligned} A\ =&\ \int_{s_0}^{s_n}(2\pi y)\ ds\\ =&\ \int_{x_0}^{x_n}2\pi y\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx\\ =&\ \int_{y_0}^{y_n}2\pi y\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy \end{aligned} $$
If $y=f(x)$ rotate about the $y$-axis:
$$ \begin{aligned} A\ =&\ \int_{s_0}^{s_n}(2\pi x)\ ds\\ =&\ \int_{y_0}^{y_n}2\pi x\sqrt{1+\left(\frac{dx}{dy}\right)^2}\ dy\\ =&\ \int_{x_0}^{x_0}2\pi x\sqrt{1+\left(\frac{dy}{dx}\right)^2}\ dx \end{aligned} $$
Example 1:
Calculate a surface area of a sphere with radius $a$.
$$ \begin{aligned} \text{Surface area }A\ =&\ \int_{-a}^{a} 2\pi\sqrt{a^2-x^2}\sqrt{1+\frac{x^2}{a^2-x^2}}\ dx\\ =&\ \int_{-a}^{a} 2\pi a\ dx\\ =&\ 2\pi a x\Big\vert_{x=-a}^{a}\\ =&\ 4\pi a^2 \end{aligned} $$
Example 2:
Calculate a surface area of a torus which is the graph of $(x-R)^2+y^2=r^2$ rotating about the $y$-axis, where $R$ and $r$ are constant, and satisfy $R\ge r$.
Consider the half part above $x$-axis of the torus, we can write the formula of $y$: $$ y\ =\ \sqrt{r^2-(x-R)^2} $$
then, $$ y^{\prime}\ =\ -\frac{x-R}{\sqrt{r^2-(x-R)^2}} $$
Write the formula of the surface area of half part of the torus: $$ \begin{aligned} A_{\text{half}}\ =&\ \int_{R-r}^{R+r}2\pi x \sqrt{1+\big(y^{\prime}\big)^2}\ dx\\ =&\ 2\pi\int_{R-r}^{R+r} x\sqrt{1+\frac{(x-R)^2}{r^2-(x-R)^2}}\ dx\\ =&\ 2\pi r\int_{R-r}^{R+r} \frac{x}{\sqrt{r^2-(x-R)^2}}\ dx \end{aligned} $$
Let $u=x-R$, then $dx=du$, we get: $$ =\ 2\pi r\int_{-r}^{r}\frac{u+R}{\sqrt{r^2-u^2}}\ du $$
As the integral is range from $-r$ to $r$, the $u$ in the numerator can be cancelled. $$ \begin{aligned} =&\ 2\pi r\int_{-r}^{r}\frac{R}{\sqrt{r^2-u^2}}\ du\\ =&\ 2\pi r R\int_{-r}^{r}\frac{du}{\sqrt{r^2-u^2}}\\ =&\ 2\pi R\int_{-r}^{r}\frac{du}{\sqrt{1-\left(\displaystyle\frac{u}{r}\right)^2}} \end{aligned} $$
Let $\displaystyle\frac{u}{r}=\sin\theta$, then $du=r\cos\theta\ d\theta$. $$ \begin{aligned} =&\ 2\pi R\int_{-\pi/2}^{\pi/2} \frac{r\cos\theta}{\sqrt{1-\sin^2\theta}}\ d\theta\\ =&\ 2\pi R r\int_{-\pi/2}^{\pi/2} \frac{d\theta}{\cos\theta}\\ =&\ 2\pi^2 R r \end{aligned} $$
Therefore, $$ \begin{aligned} A\ =&\ 2A_{\text{half}}\\ =&\ 4\pi^2 R r \end{aligned} $$