Recall that integration is the inverse operation of differentiation. Computing the integral \(\int f(x) \, dx\) is concerned with finding a function \(F\) such that \(F'(x) = f(x)\) or \(F(x) = \int f(x)\, dx\). We say that \(F\) is an anti-derivative of \(f\). Basic integrals are:
The Method of Substitution for computing integrals is rooted in the Chain Rule which is
\[
\frac{d}{dx} g(f(x)) = g'(f(x))\cdot f'(x)
\]
Hence, if we are presented with an integral \(\int h(x)\, dx\) and the integrand takes the form
\[
h(x) = g'(f(x))\cdot f'(x)
\]
then we set \(u = f(x)\), compute \(\frac{du}{dx} = f'(x)\), solve for \(f'(x) dx = du\) and make the substitution
\begin{align*}
\int h(x)\, dx &= \int g'(f(x)) f'(x)dx\\
&= \int g'(u) du\\
& = \int g'(u) du\\
&= g(u) + C\\
&= g(f(x)) + C
\end{align*}
Evaluate the integral \(\int_0^{\pi/3} \cos^3(x)\, \sin(x)\, dx\).
The integral can be written as
\[
\int (\cos(x))^3 \sin(x)\, dx
\]
In this case, we notice that if we put \(u = \cos(x)\) then \(\frac{du}{dx} = -\sin(x)\). Therefore, \(\sin(x)dx = -du\) and then by substitution
\begin{align*}
\int (\cos(x))^3 \sin(x) \, dx &= \int -u^3\, du \\
&= -\frac{1}{4} u^4 + C\\
&= - \frac{1}{4} (\cos(x))^4 + C
\end{align*}
Evaluate the integral \(\int_{1}^2 3x\sqrt{1+x^2}\, dx\).
In this case put \(u = 1+x^2\) and then \(\frac{du}{dx} = 2x\) and thus \(xdx = \frac{1}{2} du\). Now because we are changing variables from \(x\) to \(u\) and we are given a definite integral, we must change the limits of integration. When \(x=1\) then \(u = 1+(1)^2 = 2\) and when \(x=2\) then \(u=1+2^2 = 5\). Therefore,
\begin{align*}
\int_1^2 3x\sqrt{1+x^2}\, dx &= \int_2^5 \frac{3}{2} \sqrt{u} \, du \\
&= \frac{3}{2}\cdot \frac{2}{3}u^{3/2} \Big|_2^5 \\
&= 5^{3/2} - 2^{3/2}
\end{align*}
Evaluate the integral \(\int \frac{5}{1+9x^2}\, dx\).
Evaluate the integral \(\int_0^{\pi/2} e^{\sin(x)} \cos(x)\, dx\).
Evaluate the integral \(\int \sec(x)\tan(x) \sqrt{1+\sec(x)}\, dx\).
Evaluate the integral \(\int_0^{\pi/4}\tan(x)\, dx\).
Integration By Parts
Integration by parts is a sort of inverse operation of the Product Rule of differentiation. Recall that
\[
\frac{d}{dx}(f(x)g(x)) = f'(x) g(x) + f(x) g'(x)
\]
and thus integrating both sides yields
\[
f(x)g(x) = \int f'(x)g(x)\, dx + \int f(x)g'(x)\, dx
\]
The idea of integration by parts is that we are given an integral of the form
\[
\int f(x)g'(x)\, dx
\]
and it might be easier to compute \(\int f'(x)g(x)\,dx\) instead. Then
\[
\int f(x)g'(x)\, dx = f(x) g(x) - \int f'(x)g(x)\,dx.
\]
There is a formula to easily remember integration by parts. Given the integral \(\int f(x)g'(x)\, dx\) let:
\begin{align*}
u &= f(x) & dv &= g'(x)dx \\
du &= f'(x)dx & v &= g(x)
\end{align*}
Then
\[
\int u dv = uv - \int v du
\]
Evaluate \(\int x e^{2x}dx\).
Let
\begin{align*}
u &= x & dv &= e^{2x}dx\\
du &= 1\cdot dx & v &= \tfrac{1}{2}e^{2x}
\end{align*}
Then
\begin{align*}
\int x e^{2x} dx &= uv - \int v du\\
&= x \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} dx \\
&= x \frac{e^{2x}}{2} - \frac{e^{2x}}{4} + C
\end{align*}
Evaluate \(\int x^2 \ln(x) dx\).
Choose \(u=\ln(x)\) and \(dv = x^2dx\).
Evaluate \(\int_{0}^{\pi/4} x \sin(4x)\,dx\).
When doing integration by parts with an indefinite integral:
\[
\int_a^b u\, dv = uv\Big|_{a}^b - \int_a^b v\, du
\]
Let
\begin{align*}
u &= x & dv &= \sin(4x)dx\\
du &= dx & v &= -\tfrac{1}{4}\cos(4x)
\end{align*}
Then
\begin{align*}
\int_0^{\pi/4} x\sin(4x)dx &= \left(-\frac{x}{4}\cos(4x)\right)\Big|_{0}^{\pi/4} + \tfrac{1}{4}\int_0^{\pi/4}\cos(4x)\,dx\\
&= -\tfrac{\pi}{16}\cos(\pi) +\tfrac{1}{16}\sin(4x)\Big|_{0}^{\pi/4}\\
&= \tfrac{\pi}{16}
\end{align*}
In the previous example, if we had chosen
\begin{align*}
u &= \sin(4x) & dv &= x\, dx\\
du &= 4\cos(4x) & v &= x^2
\end{align*}
then we obtain
\[
uv - \int v\, du = x^2\sin(4x) - \int 4x^2\cos(4x)\,dx
\]
and the resulting integral is more difficult than the original.
Evaluate \(\int_0^{1/2}\arcsin(x)\,dx\).
Let
\begin{align*}
u &= \arcsin(x) & dv &= 1\cdot dx\\
du &= \frac{1}{\sqrt{1-x^2}} & v &= x
\end{align*}
Then
\begin{align*}
\int_0^{1/2}\arcsin(x)\,dx &= x\arcsin(x)\Big|_0^{1/2} - \int_0^{1/2} \frac{x}{\sqrt{1-x^2}}\,dx\\
&= \tfrac{1}{2}\arcsin(1/2) + \sqrt{1-x^2}\Big|_{0}^{1/2}\\
&= \frac{\pi}{12}+\sqrt{3}/2-1
\end{align*}
Evaluate \(I = \int e^{2x}\sin(3x)\,dx\).
In this example we will have to do integration by parts twice. Let
\begin{align*}
u &= e^{2x} & dv &= \sin(3x)dx\\
du &= 2e^{2x} & v &= -\frac{1}{3}\cos(3x)
\end{align*}
Then
\begin{align*}
\int e^{2x}\sin(3x)dx &= -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}\int e^{2x}\cos(3x)dx
\end{align*}
To do the second integral, let
\begin{align*}
u &= e^{2x} & dv &= \cos(3x)dx\\
du &= 2e^{2x} & v &= \frac{1}{3}\sin(3x)
\end{align*}
Then
\begin{align*}
I &= -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{3}\left(\frac{1}{3}e^{2x}\sin(3x) - \frac{2}{3} I\right)\\
&= -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{9}e^{2x}\sin(3x) - \frac{4}{9} I
\end{align*}
and solving for \(I\) yields
\[
(1+\tfrac{4}{9})I = -\frac{1}{3}e^{2x}\cos(3x) + \frac{2}{9}e^{2x}\sin(3x)
\]
or
\[
I = -\frac{3}{13}\cos(3x)e^{2x} + \frac{2}{13}\sin(3x)e^{2x} + C
\]
Other good examples:
\(\int \ln(x)dx\)
\(\int x e^{3x}dx\)
\(\int \arctan(x)dx\)
\(\int x \sec^2(x)dx\)
\(\int e^{2x}\cos(3x)dx\)
Trigonometric Integrals
The following trigonometric identities will be useful:
Factor the odd power leaving one \(\cos(x)\) and use identity \(\cos^2(x)=1-\sin^2(x)\):
\begin{align*}
\int_0^{\pi/2} \sin^2(x)\cos^5(x)\,dx &= \int_0^{\pi/2} \sin^2(x)\cos^4(x)\cos(x)dx\\
&= \int_0^{\pi/2} \sin^2(x) (1-\sin^2(x))^2 \cos(x)dx
\end{align*}
and then use substitution \(u=\sin(x)\) and \(du=\cos(x)dx\); we must change limits of integration:
\begin{align*}
\int_0^{\pi/2} \sin^2(x)\cos^5(x)\,dx &= \int_0^{\pi/2} \sin^2(x) (1-\sin^2(x))^2 \cos(x)dx \\
&= \int_0^{1} u^2 (1-u^2)^2 du\\
&= \int_0^{1} (u^2-2u^4+u^6)du\\
&= \left(\frac{u^3}{3} - 2\frac{u^5}{5} + \frac{u^7}{7}\right)\Big|_{0}^{1}\\
&= \frac{1}{3} - 2\frac{1}{5} + \frac{1}{7} = \cdots
\end{align*}
The second case we consider is when both \(n\) and \(m\) are even. In this case we need to use the half-angle formulas:
\begin{align*}
\sin^2(\theta) &= \frac{1-\cos(2\theta)}{2}\\
\cos^2(\theta) &= \frac{1+\cos(2\theta))}{2}
\end{align*}
\[
\int \tan^m(x) \sec^n(x)dx
\]
The first case we consider is when the power of \(\sec(x)\) is even.
\(\int \tan(x) \sec^4(x)dx\)
Factor even power of \(\sec(x)\) leaving one \(\sec^2(x)\) and using identity \(\sec^2(x)=1+\tan^2(x)\) for the other; then use substitution with \(u=\tan(x)\):
\begin{align*}
\int \tan(x)\sec^4(x)dx &= \int \tan(x)\sec^2(x)\sec^2(x)dx\\
&= \int \tan(x) (1+\tan^2(x)) \sec^2(x)dx\\
&= \int u (1+u^2) du\\
&= \frac{u^2}{2} + \frac{u^4}{4} + C\\
&= \frac{\tan^2(x)}{2} + \frac{\tan^4(x))}{4} + C
\end{align*}
The second case we consider is when both powers are odd.
\(\int \tan^3(x) \sec^7(x)dx\)
Factor out one \(\tan(x)\) and one \(\sec(x)\) term; use identity \(\tan^2(x) = \sec^2(x) -1\); then use substitution \(u=\sec(x)\):
\begin{align*}
\int \tan^3(x) &\sec^7(x)dx \\
&= \int \tan^2(x)\sec^6(x) \sec(x)\tan(x)dx\\[2ex]
&= \int (\sec^2(x)-1)\sec^6(x) \sec(x)\tan(x)dx\\[2ex]
&= \int (u^2-1)u^6 du\\[2ex]
&= \frac{u^9}{9} + \frac{u^7}{7} + C\\[2ex]
&= \frac{\sec^9(x))}{9} + \frac{\sec^7(x)}{7} + C
\end{align*}
\(\int \tan^2(x)dx\)
Compute
\begin{align*}
\int\tan^2(x)dx &= \int (\sec^2(x)-1)dx \\
&= \int\sec^2(x) - \int dx\\
&= \tan(x) - x + C
\end{align*}
Other examples:
\(\int_0^{\pi/2} \sin^2(x)dx\)
\(\int \sin^3(x)\cos^3(x)dx\)
\(\int \sec^4(x)\tan^2(x)dx\)
\(\int x\sin^2(x)dx\)
Trigonometric Substitution
Trigonometric substitution is useful when the integral involves expressions of the form
\[
\sqrt{a^2-x^2}, \sqrt{a^2+x^2}, \sqrt{x^2-a^2}
\]
including
\[
(x^2-a^2)^{3/2}, \frac{1}{x^2\sqrt{a^2+x^2}}
\]
When these expressions appear the following substitutions are useful:
Expression
Substitution
\(\sqrt{a^2-x^2}\)
\(x=a\sin(\theta)\)
\(\sqrt{a^2+x^2}\)
\(x=a\tan(\theta)\)
\(\sqrt{x^2-a^2}\)
\(x=a\sec(\theta)\)
Trigonometric Substitutions
Evaluate \(\int x^3 \sqrt{9-x^2} dx\).
Let \(x=3\sin(\theta)\) and then \(dx = 3\cos(\theta)d\theta\) and from \(\frac{x}{3}=\sin(\theta)\) the triangle is
Now make the substitutions:
\begin{align*}
x &= 3\sin(\theta)\\
dx &= 3\cos(\theta)d\theta
\end{align*}
and evaluate
\begin{align*}
\int x^3 \sqrt{9-x^2} dx &= \int 3^3\sin^3(\theta)\sqrt{9-9\sin^2(\theta)} 3\cos(\theta)d\theta \\
&= 3^5\int \sin^3(\theta) \cos^2(\theta)d\theta\\
&= 3^5 \int (1-\cos^2(\theta))\cos^2(\theta) \sin(\theta)d\theta\\
&= -3^5 \int (1-u^2)u^2\, du \qquad u=\cos(\theta)\\
&= -3^5 \left( \tfrac{1}{3}u^3 - \tfrac{1}{5} u^5 \right) + C\\
&= -\frac{3^5}{3} (\cos(\theta))^3 + \frac{3^5}{5} (\cos(\theta))^5 + C \\
&= -\frac{3^5}{3} \left(\frac{\sqrt{9-x^2}}{3}\right)^3 + \frac{3^5}{5} \left(\frac{\sqrt{9-x^2}}{3}\right)^5 + C
\end{align*}
Evaluate \(\int \frac{1}{\sqrt{4+x^2}} dx\).
Let \(x=2\tan(\theta)\) and then \(dx = 2\sec^2(\theta)d\theta\), and from the relationship \(\frac{x}{2}=\tan(\theta)\) the triangle is:
Now make the substitutions
\begin{align*}
x&=2\tan(\theta)\\
dx&=2\sec^2(\theta)d\theta
\end{align*}
and evaluate:
\begin{align*}
\int \frac{1}{\sqrt{4+x^2}} \, dx &= \int\frac{2\sec^2(\theta)}{\sqrt{4+4\tan^2(\theta)}} \, d\theta\\
&= \int \frac{2\sec^2(\theta)}{\sqrt{4\sec^2(\theta)}} \, d\theta \\
&= \int 2\sec(\theta)\, d\theta \\
&= \ln |\sec(\theta) + \tan(\theta)| + C
\end{align*}
Using the triangle we have
\[
\sec(\theta) = \frac{\sqrt{4+x^2}}{2}
\]
and \(\tan(\theta) = x/2\). Therefore,
\begin{align*}
\int \frac{1}{\sqrt{4+x^2}} \, dx &= \ln |\sec(\theta) + \tan(\theta)| + C \\
&= \ln\left| \frac{\sqrt{4+x^2}}{2} + \frac{x}{2}\right| + C
\end{align*}
Let \(x=\sqrt{7}\sin(\theta)\), then \(dx=\sqrt{7}\cos(\theta)\), and
Then
\begin{align*}
\int \frac{x^2}{(7-x^2)^{3/2}} \, dx &= \int \frac{7\sin^2(\theta) \sqrt{7}\cos(\theta)}{ (7-7\sin^2(\theta))^{3/2}} \,d\theta \\
&= \int \frac{7\sin^2(\theta) \sqrt{7}\cos(\theta)}{ 7^{3/2} (\cos^2(\theta))^{3/2}} \,d\theta \\
&= \int \frac{\sin^2(\theta)}{\cos^2(\theta)}\,d\theta \\
&= \int \tan^2(\theta) \, d\theta\\
&= \int (\sec^2(\theta) - 1) d\theta\\
&= \tan(\theta) - \theta + C
\end{align*}
From the triangle:
\[
\tan(\theta) = \frac{x}{\sqrt{7-x^2}}
\]
and from \(x=\sqrt{7}\sin(\theta)\) we obtain \(\theta = \arcsin(x/\sqrt{7})\). Therefore,
\begin{align*}
\int \frac{x^2}{(7-x^2)^{3/2}} \, dx &= \tan(\theta) - \theta + C\\
& = \frac{x}{\sqrt{7-x^2}} - \arcsin(x/\sqrt{7}) + C
\end{align*}
Evaluate \(\int \frac{3}{\sqrt{4+9x^2}}\, dx\).
Notice the \(9\) in front of \(x^2\). We can factor it out:
\[
\int \frac{3}{\sqrt{4+9x^2}}\, dx = \int \frac{3}{9\sqrt{\tfrac{4}{9} + x^2}}\, dx
\]
Then let \(x=\sqrt{\tfrac{4}{9}}\tan(\theta) = \tfrac{2}{3} \tan(\theta)\) and then
\[
dx = \tfrac{2}{3}\sec^2(\theta) \, d\theta
\]
Using \(3x/2 = \tan(\theta)\), the triangle is
Then
\begin{align*}
\int \frac{3}{\sqrt{4+9x^2}}\, dx &= \int \frac{2\sec^2(\theta)}{\sqrt{4+4\tan^2(\theta)}}\,d\theta \\
&= \int \sec(\theta) d\theta\\
&= \ln|\sec(\theta) + \tan(\theta)| + C\\
&= \ln\left|\frac{\sqrt{4+9x^2}}{2} + \frac{3x}{2}\right| + C
\end{align*}
Evaluate \(\int\frac{\sqrt{x^2-25}}{x^3}dx\).
Let
\begin{align*}
x &= 5\sec(\theta)\\
dx &= 5\sec(\theta)\tan(\theta)
\end{align*}
and the triangle is
Then
\begin{align*}
\int\frac{\sqrt{x^2-25}}{x^3}dx &= \int \frac{\sqrt{25\sec^2(\theta) - 25}}{125\sec^3(\theta)} 5 \sec(\theta)\tan(\theta)d\theta\\
&= \int \frac{\tan^2(\theta)}{5\sec^2(\theta)}d\theta\\
&= \frac{1}{5} \int\sin^2(\theta)d\theta\\
&= \frac{1}{5} \int \frac{1-\cos(2\theta)}{2}d\theta\\
&= \frac{\theta}{10} - \frac{1}{20} \sin(2\theta) + C\\
&= \frac{1}{10} \arcsec(x/5) - \frac{1}{10}\sin(\theta) \cos(\theta) + C\\
&= \frac{1}{10} \arcsec(x/5) - \frac{\sqrt{x^2-25}}{2x^2} + C
\end{align*}
Evaluate \(\int x^3 \sqrt{1+x^2}dx\).
Let \(x=\tan(\theta)\) and then \(dx=\sec^2(\theta)d\theta\), and the triangle is
Then
\begin{align*}
\int x^3 \sqrt{1+x^2}dx &= \int \tan^3(\theta) \sqrt{1+\tan^2(\theta)} \sec^2(\theta)d\theta\\
&= \int \tan^3(\theta)\sec^3(\theta) d\theta\\
&= \int \tan^2(\theta)\sec^2(\theta)\sec(\theta)\tan(\theta)d\theta\\
&= \int (u^2-1)u^2 du \qquad u=\sec(\theta)\\
&= \tfrac{1}{5} u^5 - \tfrac{1}{3} u^3 + C\\[2ex]
&= \tfrac{1}{5} (\sec(\theta))^5 - \tfrac{1}{3} (\sec(\theta))^3 + C\\[2ex]
&= \tfrac{1}{5} (\sqrt{1+x^2})^5 - \tfrac{1}{3} ( \sqrt{1+x^2})^3 + C
\end{align*}
Evaluate \(\int \frac{x}{25+x^2}dx\).
It is easier to use the substitution \(u=25+x^2\) then to use trigonometric substitution.
Partial Fractions
The method of partial fraction is suitable for integrals of rational functions:
\[
\int \frac{P(x)}{Q(x)}dx
\]
where both \(P\) and \(Q\) are polynomials. For example,
\[
\int \frac{x+5}{x^2+x-2}\, dx, \qquad \int \frac{x^3-x+1}{x^2 + 9}\, dx
\]
We say that \(f(x) = \frac{P(x)}{Q(x)}\) is a proper rational function if \(\deg(P) \lt \deg(Q)\). For example, the following are proper rational functions:
\[
\frac{x+5}{x^2+x-2}, \qquad \frac{x^4-x^2+7}{3x^7 - 7}
\]
whereas the following are improper rational functions:
\[
\frac{x^2-x+1}{3x^2+7}, \qquad \frac{x^3-x+1}{x^2 + 9}
\]
The method of partial fractions is really about decomposing a rational function \(f(x) = \frac{P(x)}{Q(x)}\) into a sum of simpler rational functions. For example,
\[
\frac{x+5}{x^2+x-2} = \frac{x+5}{(x+2)(x-1)}
\]
and one can verify that
\[
\frac{x+5}{(x+2)(x-1)} = \frac{-1}{x+2} + \frac{2}{x-1}
\]
Then
\begin{align*}
\int \frac{x+5}{x^2+x-2}dx &= \int \frac{x+5}{(x+2)(x-1)}dx\\
&= \int \left( \frac{-1}{x+2} + \frac{2}{x-1}\right) dx\\
&= \int \frac{-1}{x+2}dx + \int\frac{2}{x-1}dx \\
&= -\ln|x+2| + 2\ln|x-1| + C
\end{align*}
We now discuss the general method; we first only consider proper rational functions, and later we will deal with the improper case. In all cases we consider, the starting point is to factor the denominator \(Q(x)\) in \(P(x)/Q(x)\).\\
\(Q(x)\) factors into distinct linear terms
Evaluate \(\int \frac{3}{x^2+x-6}dx\).
The denominator factors as \[x^2+x-6 = (x+3)(x-2).\] We then seek the decomposition:
\[
\frac{3}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2}
\]
where \(A\) and \(B\) are to be determined. Now
\[
\frac{A}{x+3} + \frac{B}{x-2} = \frac{A(x-2) + B(x+3)}{(x+3)(x-2)} = \frac{(A+B)x + 3B-2A}{(x+3)(x-2)}
\]
Now we equate numerators
\[
3 = (A+B)x + 3B-2A
\]
equating coefficients yield the equations:
\begin{align*}
A+B &=0 \\
3B-2A &= 3
\end{align*}
Solving for \(A\) and \(B\) we obtain \(A=-3/5\) and \(B=3/5\). Therefore, the partial fraction decomposition is
\[
\frac{3}{(x+3)(x-2)} = \frac{(-3/5)}{x+3} + \frac{(3/5)}{x-2}
\]
and then
\begin{align*}
\int \frac{3}{x^2+x-6}dx &= \int \frac{3}{(x+3)(x-2)} dx \\[2ex]
&=\int \left(\frac{(-3/5)}{x+3} + \frac{(3/5)}{x-2}\right)dx \\[2ex]
&= \frac{-3}{5} \ln|x+3| + \frac{3}{5} \ln|x-2| + C
\end{align*}
There is an alternative and quick way to find \(A\) and \(B\). From the relationship
\[
\frac{3}{(x+3)(x-2)} = \frac{A}{x+3} + \frac{B}{x-2} = \frac{A(x-2) + B(x+3)}{(x+3)(x-2)}
\]
and equating the numerators we obtain
\[
3 = A(x-2) + B(x+3).
\]
Evaluating at \(x=2\) yields \(3=5B\) and thus \(B=3/5\), and evaluating at \(x=-3\) yields \(3=-5A\) and thus \(A=-3/5\).
Write
\begin{align*}
\frac{3x^2-2}{(x-1)(x-2)(x+1)} &= \frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x+1}
\end{align*}
and then
\begin{gather*}
\frac{A}{x-1} + \frac{B}{x-2} + \frac{C}{x+1} \\[2ex] = \frac{A(x-2)(x+1) + B(x-1)(x+1) + C(x-1)(x-2)}{(x-1)(x-2)(x+1)}
\end{gather*}
Equating the numerators we obtain
\[
3x^2-2 = A(x-2)(x+1) + B(x-1)(x+1) + C(x-1)(x-2)
\]
Evaluating both sides at \(x=1\) yields \(1=-2A\) and thus \(A=-1/2\), evaluating at \(x=-1\) yields \(1 = 6C\) and thus \(C=1/6\), and evaluating at \(x=2\) yields \(10 = 3B\) and thus \(B=10/3\). Thus
\begin{align*}
\int \frac{3x^2-2}{(x-1)(x-2)(x+1)}dx &= \int \left(\frac{-1/2}{x-1} + \frac{10/3}{x-2} + \frac{1/6}{x+1} \right)dx \\[2ex]
&= -\frac{1}{2}\ln|x-1| + \frac{10}{3} \ln|x-2| \\
&\qquad + \frac{1}{6} \ln|x+1| + C
\end{align*}
In general, if \(Q(x)\) factors as
\[
Q(x) = (x+a)(x+b)\cdots (x+c)
\]
then we seek the decomposition
\[
\frac{P(x)}{Q(x)} = \frac{A}{x+a} + \frac{B}{x+b} + \cdots + \frac{C}{x+c}
\]
\(Q(x)\) factors into linear terms some of which are repeated
If the denominator \(Q(x)\) contains a repeated factor, such as \((x+2)^4\), then in the partial fraction decomposition we include the terms
\[
\frac{A}{x+2} + \frac{B}{(x+2)^2} + \frac{C}{(x+2)^3} + \frac{D}{(x+2)^4}
\]
Evaluate \(\int \frac{4}{(x+3)(x-1)^2}dx\).
We seek the decomposition
\begin{align*}
\frac{4}{(x+3)(x-1)^2} &= \frac{A}{x+3} + \frac{B}{(x-1)} + \frac{C}{(x-1)^2}\\
&= \frac{A(x-1)^2 + B(x-1)(x+3) + C(x+3)}{(x+3)(x-1)^2}
\end{align*}
and equating numerators yields
\[
4 = A(x-1)^2 + B(x-1)(x+3) + C(x+3)
\]
Evaluating at \(x=1\) yields \(4=4C\) and thus \(C=1\) and evaluating at \(x=-3\) yields
\[
4 = 16A
\]
and thus \(A=1/4\). To obtain another equation we could evaluate at another convenient value of \(x\), say at \(x=0\):
\[
4 = A - 3B + 3C = \frac{1}{4} - 3B + 3
\]
and solving for \(B\) yields \(B=-\frac{1}{4}\). Therefore,
\begin{align*}
\int \frac{4}{(x+3)(x-1)^2}dx &= \int \frac{1/4}{x+3} + \frac{-1/4}{(x-1)} + \frac{1}{(x-1)^2}\, dx\\[2ex]
&= \frac{1}{4}\ln|x+3| -\frac{1}{4}\ln|x-1| - \frac{1}{x-1} + C
\end{align*}
Write the form of the partial fraction decomposition for
\[
f(x) = \frac{x^2+3}{(x+1)^2(x-2)^3(x+7)}.
\]
The partial fraction decomposition takes the form
\begin{align*}
f(x) &= \frac{A}{x+1}+\frac{B}{(x+1)^2} + \frac{C}{(x-2)} + \frac{D}{(x-2)^2}+ \frac{E}{(x+7)}
\end{align*}
\(Q(x)\) contains irreducible quadratic terms, none are repeated
An irreducible quadratic polynomial is a quadratic polynomials whose roots are complex numbers. For example, \(x^2+1\) or \(x^2-2x+3\), and you can verify that the roots are complex because the discriminant \(b^2 - 4ac\) is negative. If \(Q(x)\) has the irreducible term \(ax^2+bx+c\) we introduce in the partial fraction decomposition the term
\[
\frac{Ax + B}{ax^2+bx+c}
\]
Evaluate \(\int \frac{x^2+2}{(x+1)(x^2+1)}dx\).
The partial fraction decomposition takes the form
\begin{align*}
\frac{x^2+2}{(x+1)(x^2+1)} &= \frac{A}{x+1} + \frac{Bx+C}{x^2+1} \\
&= \frac{A(x^2+1) + (x+1)(Bx+C)}{(x+1)(x^2+1)}
\end{align*}
and equating numerators we obtain
\[
x^2 + 2 = A(x^2+1) + (x+1) (Bx+C)
\]
Evaluating both sides at \(x=-1\) yields \(3 = 2A\) and thus \(A=3/2\). Evaluating at \(x=0\) and \(x=1\) yield
\begin{align*}
2 &= A + C\\
3 &= 2A + 2(B+C)
\end{align*}
and thus \(C=2-A = 1/2\) and solving for \(B\) yields \(B=-1/2\). Therefore,
\begin{align*}
\int \frac{x^2+2}{(x+1)(x^2+1)}dx &= \int \left(\frac{3/2}{x+1} + \frac{-\tfrac{1}{2}x + \tfrac{1}{2}}{x^2+1}\right) dx\\
&= \tfrac{3}{2}\ln|x+1| - \tfrac{1}{4}\ln|x^2+1| + \tfrac{1}{2}\arctan(x)
\end{align*}
Now we consider the case when \(P(x)/Q(x)\) is improper. Every improper rational function can be written as
\[
\frac{P(x)}{Q(x)} = \underbrace{g(x)}_{\text{polynomial}} + \underbrace{\frac{p(x)}{Q(x)}}_{\text{proper rational}}
\]
This can be accomplished using long division of polynomials.
Compute
\[
\int \frac{x^3-2x^2-4}{x-3}\, dx.
\]
The rational \(\frac{x^3-2x^2-4}{x-3}\) is improper and thus we must use long division which yields
\[
\frac{x^3-2x^2-4}{x-3} = x^2 + x + 3 + \frac{5}{x-3}
\]
Therefore,
\begin{align*}
\int \frac{x^3-2x^2-4}{x-3}\, dx & = \int \left(x^2 + x + 3 + \frac{5}{x-3} \right)\, dx\\
&= \frac{1}{3}x^3 + \frac{1}{2}x^2 + 3x + 5 \ln|x-3| + C.
\end{align*}
Compute
\[
\int \frac{x^3-9}{x^2+4x+3} \, dx
\]
After performing long division one obtains that
\[
\frac{x^3-9}{x^2+4x+3} = x-4 + \frac{13x+3}{x^2+4x+3}
\]
Therefore,
\[
\int \frac{x^3-9}{x^2+4x+3} \, dx = \int (x-4)\, dx + \int \frac{13x+3}{x^2+4x+3}dx
\]
Use partial fraction decomposition we get
\[
\frac{13x+3}{x^2+4x+3} = \frac{18}{x+3} + \frac{-5}{x+1}
\]
Hence
\[
\int \frac{x^3-9}{x^2+4x+3} \, dx = x^4 -4x + 18\ln|x+3| -5\ln|x+1|+C
\]
Numerical Integration
In many cases, it's not possible to compute an integral in closed-form. Examples of such integrals are
\[
\int e^{-x^2}dx, \int \sqrt{1+x^3}\, dx, \int\sqrt{1+\cos^2(x)}\, dx.
\]
It isn't that these integrals do not exist but rather they cannot be expressed in terms of familiar functions. However, for example, we may want to compute the definite integral
\[
\int_0^2 e^{-x^2}\, dx
\]
which geometrically corresponds to the area under the graph of \(f(x) = e^{-x^2}\) from \(x=0\) to \(x=2\). The best we can hope for is to approximate the definite integral. You are already familiar with approximating definite integrals using Riemann sums. Recall that Riemann sums are used to approximate a definite integral
\[
\int_a^b f(x) \, dx.
\]
In the above graph, a hypothetical function \(f\) is graphed from \(x=a\) to \(x=b\), here \(a=-2\) and \(b=4\). A Riemann sum is simply the sum of the areas of the rectangles and is an approximation to the area between the graph and the \(x\)-axis. To build a Riemann sum, we partition the interval \([a,b]\) into \(n\) equal subintervals; in the above picture \(n=12\) (notice there are 12 rectangles). The width of each rectangle is then
\[
\Delta x = \frac{b-a}{n} = \frac{4-(-2)}{12} = \frac{1}{2}.
\]
The partition of \([a,b]=[-2,4]\) is
In each sub-interval \([x_{k-1}, x_k]\) we pick some point \(c_k\) and then the height of the rectangle over the sub-interval \([x_{k-1},x_k]\) is \(f(c_k)\). Thus the area of the rectangle over the interval \([x_k,x_{k-1}]\) is \(f(c_k) \Delta x\). We then add up all the areas and this is a Riemann sum
\[
\sum_{k=1}^n f(c_k)\Delta x = \Delta x (f(c_1) + f(c_2) + f(c_3) + \cdots + f(c_n))
\]
In Calculus I, the point \(c_k\) was usually chosen as the mid-point, left-end point, or right-end point of the interval \([x_{k-1}, x_k]\). In any of those cases, what we are doing is approximating the function \(f\) on the interval \([x_{k-1},x_k]\) with the constant value \(f(c_k)\).
Use a Riemann sum with \(n=4\) sub-intervals to estimate \(\int_0^{2\pi} \sqrt{2+2\sin(x)}dx\). Use the right-end points of each sub-interval.
We have \(\Delta x = \frac{2\pi-0}{4} = \frac{\pi}{2}\). The end-points of the sub-intervals are
\(x_0\)
\(x_1\)
\(x_2\)
\(x_3\)
\(x_4\)
0
\(\frac{\pi}{2}\)
\(\pi\)
\(\frac{3\pi}{2}\)
\(2\pi\)
Partition of \([0,2\pi]\)
Hence, the intervals are \([0,\tfrac{\pi}{2}]\), \([\tfrac{\pi}{2},\pi]\), \([\pi,\tfrac{3\pi}{2}]\), and \([\tfrac{3\pi}{2},2\pi]\). Using right-end points for each sub-interval, the Riemann sum is
\begin{align*}
\sum_{k=1}^4 f(c_k) \Delta x &= \Delta x \left( f(\tfrac{\pi}{2}) + f(\pi) + f(\tfrac{3\pi}{2}) + f(2\pi) \right) \\
&= \frac{\pi}{2} \left(\sqrt{4} + \sqrt{2} + \sqrt{0} + \sqrt{2}\right)\\
&= \pi(1+\sqrt{2})
\end{align*}
Therefore,
\[
\int_0^{2\pi}\sqrt{2+2\sin(x)}dx \approx \pi (1+\sqrt{2})
\]
Trapezoidal Rule
When using the mid-point, right-end point, or the left-end point rule in a Riemann sum, we are approximating the \(f\) on each subinterval \([x_{k-1}, x_k]\) with the constant value \(f(c_k)\). In the Trapezoidal rule, we approximate \(f\) on \([x_{k-1}, x_k]\) with a line from \((x_{k-1}, f(x_{k-1}))\) to \((x_k, f(x_k))\):
Notice that the area under the line is the area of a trapezoid, which is
\[
\frac{1}{2}\big[f(x_{k-1}) + f(x_k) \big] \Delta x
\]
Approximating the area between the graph of \(f\) and the \(x\)-axis over each sub-interval with the areas of trapezoids results in the following picture:
Adding up all the areas of the trapezoids and simplifying we obtain the Trapezoidal rule
\[
T = \frac{\Delta x}{2}\left[f(x_0)+2f(x_1)+2f(x_2)+\cdots+2f(x_{n-1}) + f(x_n)\right]
\]
The number \(T\) obtained by the trapezoidal rule is the average of the right-end point and left-end point approximations.
Use the trapezoidal rule to approximate the definite integral
\[
\int_{-1}^1 \frac{x^2}{1+x^2}\, dx
\]
Use \(n=6\) subintervals.
The form of the trapezoidal rule takes the form
\begin{align*}
T &= \frac{\Delta x}{2}\big[f(x_0)+2f(x_1)+2f(x_2)+2f(x_3)+2f(x_4) \\
&\hspace{2cm}+2f(x_5)+f(x_6)\big]
\end{align*}
First compute
\[
\Delta x = \frac{b-a}{n} = \frac{1-(-1)}{6} = \frac{1}{3}
\]
Now compute the grid points \(x_k\) and \(f(x_k)\):
\begin{align*}
x_0 &= -1 & f(x_0) &= 1/2\\
x_1 &= -2/3 & f(x_1) &= 4/13\\
x_2 &= -1/3 & f(x_2) &= 1/10\\
x_3 &= 0 & f(x_3) &= 0\\
x_4 &= 1/3 & f(x_4) &= 1/10\\
x_5 &= 2/3 & f(x_5) &= 4/13\\
x_6 &= 1 & f(x_6) &= 1/2
\end{align*}
Then
\begin{align*}
T &= \frac{(1/3)}{2}\left[\tfrac{1}{2}+2\tfrac{4}{13}+2\tfrac{1}{10}+2\cdot 0 + 2\tfrac{1}{10}+2\tfrac{4}{13} +\tfrac{1}{2}\right]\\
&= \frac{59}{195}
\end{align*}
Simpson's Rule
In Simpson's rule, we approximate \(f\) over two subintervals with a parabola:
The area of the parabola (red curve) that passes through the points \((x_0, f(x_0))\), \((x_1,f(x_1))\), and \((x_2,f(x_2))\) is
\[
\frac{\Delta x}{3} \left[f(x_0)+4f(x_1)+f(x_2)\right]
\]
where \(\Delta x\) is the step-size between \(x_0,x_1,x_2\). If we repeat this procedure we obtain the following picture in the case that \(n=8\):
Because we need two sub-intervals for each parabola, we always need \(n\) to be even for Simpson's rule. Adding up all the areas under the parabolas and simplifying we obtain Simpson's rule:
\begin{align*}
S &= \frac{\Delta x}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4)\\
&\hspace{1cm} \cdots + 2f(x_{n-2})+4f(x_{n-1}) + f(x_n)]
\end{align*}
Use Simpson's rule to approximate the definite integral
\[
\int_{-1}^{1} \frac{x^2}{1+x^2}dx
\]
Use \(n=8\) subintervals.
With \(n=8\), Simpson's rule takes the form
\begin{align*}
S &= \frac{\Delta x}{3}[f(x_0)+4f(x_1)+2f(x_2)+4f(x_3)+2f(x_4) \\
&\hspace{1cm} 4f(x_5) + 2f(x_6) + 4f(x_7) + f(x_8)]
\end{align*}
In this case
\[
\Delta x = \frac{b-a}{n} = \frac{1 - (-1)}{8} = \frac{1}{4}
\]
The grid points and the values of \(f\) at these points are:
\begin{align*}
x_0 &= -1 & f(x_0) &= 1/2\\
x_1 &= -3/4 & f(x_1) &= 9/25\\
x_2 &= -1/2 & f(x_2) &= 1/5\\
x_3 &= -1/4 & f(x_3) &= 1/17\\
x_4 &= 0 & f(x_4) &= 0\\
x_5 &= 1/4 & f(x_5) &= 1/17\\
x_6 &= 1/2& f(x_6) &= 1/5\\
x_7 &= 3/4 & f(x_7) &= 9/25\\
x_8 &= 1 & f(x_8) &= 1/2
\end{align*}
Therefore,
\[
S = \frac{\Delta x}{3} \big[\tfrac{1}{2}+4\tfrac{9}{25}+2\tfrac{1}{5}+4\tfrac{1}{17}+2\cdot 0+4\tfrac{1}{17}+2\tfrac{1}{5}+4\tfrac{9}{25}+\tfrac{1}{2}\big] = \tfrac{2189}{5100}
\]
Improper Integrals
The first type of improper integrals that we consider are one with an infinite interval of integration. For example, consider
\[
\int_1^\infty \frac{1}{x^2}\,dx
\]
Interpreting an integral as the area between the graph of the integrand \(f(x)=\frac{1}{x^2}\) and the \(x\)-axis, then the above integral represents the area, possibly infinite, shown below:
If the value of the integral \(\int_0^\infty \frac{1}{x^2}dx\) is finite then we say that the improper integral converges otherwise we say that it diverges.
Evaluate \(\int_1^\infty \frac{1}{x^2}dx\).
We first consider
\begin{align*}
\int_1^t \frac{1}{x^2} dx &= \frac{-1}{x}\Big|_1^t \\
&= -\frac{1}{t} + 1
\end{align*}
Then
\begin{align*}
\int_1^\infty \frac{1}{x^2} dx &= \lim_{t\rightarrow\infty}\left( \int_0^t \frac{1}{x^2} dx\right)\\
&= \lim_{t\rightarrow\infty} \left(\frac{-1}{t} + 1\right)\\
&= 1.
\end{align*}
Therefore, since the value of the integral is finite we say that the improper integral converges.
Evaluate \(\int_{-\infty}^0 e^{3x}dx\).
We first consider
\[
\int_t^0 e^{3x} dx = \frac{1}{3}e^{3x}\Big|_t^0 = \frac{1}{3}e^0 - \frac{1}{3}e^{3t}=\frac{1}{3}-\frac{1}{3}e^{3t}
\]
Therefore,
\begin{align*}
\int_{-\infty}^0 e^{3x} dx &= \lim_{t\rightarrow-\infty} \left(\int_t^0 e^{3x} dx\right)\\
&= \lim_{t\rightarrow-\infty} \left(\frac{1}{3}-\frac{1}{3}e^{3t}\right)\\
&= \frac{1}{3} - \frac{1}{3}\lim_{t\rightarrow-\infty} e^{3t}\\
&= \frac{1}{3}
\end{align*}
Therefore, the improper integral converges to \(\frac{1}{3}\).
Evaluate \(\int_1^{\infty} \frac{1}{x}dx\).
We compute
\begin{align*}
\int_1^{\infty} \frac{1}{x}dx &= \lim_{t\rightarrow\infty}\left( \int_1^t \frac{1}{x}dx\right)\\
&=\lim_{t\rightarrow\infty}\left(\ln|x|\Big|_1^t\right)\\
&= \lim_{t\rightarrow\infty}\left(\ln|t| - \ln|1|\right)\\
&= \lim_{t\rightarrow\infty}\left(\ln|t|\right)\\
&= \infty
\end{align*}
Hence, the improper integral diverges.
Evaluate \(\int_{-\infty}^\infty x^2e^{x^3}dx\).
We need to break up the interval of integration first (choose \(x=0\) as the break-up point, any can be chosen):
\[
\int_{-\infty}^\infty x^2e^{x^3}dx = \int_{-\infty}^0 x^2 e^{x^3}dx + \int_{0}^\infty x^2 e^{x^3}dx
\]
We do each one separately and if even one diverges then the whole integral diverges. We first compute the anti-derivative using the substitution \(u=x^3\), \(du=3x^2dx\):
\[
\int x^2 e^{x^3}dx = \int \frac{1}{3}e^u du = \frac{1}{3}e^{x^3}.
\]
Then
\begin{align*}
\int_{-\infty}^0 x^2 e^{x^3}dx &= \lim_{t\rightarrow-\infty} \int_t^0 x^2e^{x^3}dx\\
&= \lim_{t\rightarrow-\infty} \left( \frac{1}{3}e^{x^3}\right)\Big|_{t}^0\\
&= \lim_{t\rightarrow-\infty} \left(\frac{1}{3}e^{0} - \frac{1}{3}e^{t^3}\right)\\
&= \frac{1}{3}
\end{align*}
Now do the other integral:
\begin{align*}
\int_{0}^\infty x^2 e^{x^3}dx &= \lim_{t\rightarrow\infty} \int_0^t x^2e^{x^3}dx\\
&= \lim_{t\rightarrow\infty} \left( \frac{1}{3}e^{x^3}\right)\Big|_{0}^t\\
&= \lim_{t\rightarrow\infty} \left(\frac{1}{3}e^{t^3}-\frac{1}{3}e^0\right)\\
&= \infty
\end{align*}
Thus, one half of the improper integral diverges and thus the entire improper integral diverges.
We now consider improper integrals whose integrand has a singularity (division by zero) in the interval of integration. For example, consider
\[
\int_0^1 \frac{1}{x^{1/3}}dx
\]
Notice that the integrand \(f(x)=\frac{1}{x^{1/3}}\) has a singularity at \(x=0\) which is indeed inside the interval \([0,1]\) of integration. At \(x=0\) the function \(f\) has a vertical asymptote. In this case, we evaluate the integral
\[
\int_t^1 \frac{1}{x^{1/3}}dx = \frac{x^{2/3}}{2/3}\Big|_{t}^1 = \frac{3}{2} - \frac{3}{2}t^{2/3}
\]
and then compute the one-sided limit
\[
\lim_{t\rightarrow 0^+} \left( \int_t^1 \frac{1}{x^{1/3}}dx \right) = \lim_{t\rightarrow 0^+} \left(\frac{3}{2} - \frac{3}{2}t^{2/3}\right) = \frac{3}{2}.
\]
We then say that the improper integral converges and it converges to
\[
\int_0^1 \frac{1}{x^{1/3}}dx = \frac{3}{2}.
\]
Evaluate \(\int_2^4 \frac{1}{(x-4)^3}dx\).
The integrand has a vertical asymptote at \(x=4\). We first compute
\begin{align*}
\int_2^t \frac{1}{(x-4)^3}dx &= \frac{(x-4)^{-2}}{-2}\Big|_{2}^t \\
&= \frac{-1}{2(t-4)^2} + \frac{1}{8}
\end{align*}
and then we compute the one-sided limit:
\begin{align*}
\lim_{t\rightarrow 4^{-}} \left(\int_2^t \frac{1}{(x-4)^3}dx\right) &= \lim_{t\rightarrow 4^{-}} \left( \frac{-1}{2(t-4)^2} + \frac{1}{8}\right) \\
&= -\infty
\end{align*}
Thus the improper integral diverges.
Evaluate \(\int_0^9 \frac{1}{(x-5)^2}dx\).
In this case, the integrand has a singularity at the point \(x=5\) which is not a boundary point of the interval \([0,9]\). We therefore need to break-up the integral as follows:
\begin{align*}
\int_0^9 \frac{1}{(x-5)^2}dx = \int_0^5 \frac{1}{(x-5)^2}dx +\int_5^9 \frac{1}{(x-5)^2}dx
\end{align*}
and then we do each one separately. We compute
\begin{align*}
\int_0^5 \frac{4}{(x-5)^2}dx &= \lim_{t\rightarrow 5^{-}} \left( \int_0^t \frac{4}{(x-5)^2}dx \right)\\
&= \lim_{t\rightarrow 5^{-}} \left(\frac{-4}{(x-5)}\right)\Big|_0^t\\
&= \lim_{t\rightarrow 5^{-}} \left(\frac{-4}{(t-5)} - \frac{4}{5}\right)\\
&= \infty
\end{align*}
Thus, one-half of the improper integral diverges and thus the entire improper integral diverges.
