The Art and Practice
of Calculating Integrals
-
First, before writing anything down,
look over each of the following integrals
and decide if doing a substitution
or if taking the integral by parts
is a better first idea.
Note that there is no correct decision —
maybe one decision is better,
and maybe neither option works very well.
Consider whether knowing a previous integral helps.
Then after you’ve looked over all of them,
try evaluating each one
to see if your choice was a good one,
or if the integral is tougher than it appears.
\[
\int \sin(x)\cos(x) \,\mathrm{d}x
\qquad
\int x\sin(x)\cos(x) \,\mathrm{d}x
\qquad
\int x\sin\!\left(x^2\right)\cos\!\left(x^2\right) \,\mathrm{d}x
\]
\[
\int x\cos\!\left(x\right) \,\mathrm{d}x
\qquad
\int x\cos\!\left(x^2\right) \,\mathrm{d}x
\qquad % not expressible in terms of elementary functions
\int x^2\cos\!\left(x^2\right) \,\mathrm{d}x
\qquad
\int x^3\cos\!\left(x^2\right) \,\mathrm{d}x
\]
\[
\int \frac{1}{x}\cos\!\left(\ln(x)\right) \,\mathrm{d}x
\qquad
\int x\ln(x) \,\mathrm{d}x
\qquad
\int \cos\!\left(\ln(x)\right) \,\mathrm{d}x
\qquad
\int x\cos\!\left(\ln(x)\right) \,\mathrm{d}x
\]
-
Write down a formula for the integral
\[ \int x^n\cos(x) \,\mathrm{d}x \]
where \(n\) is an arbitrary positive integer.
If this task feels overwhelming, start computing the integral
for some specific large value of \(n\), say \(n=10\),
and see if you notice the pattern.
If you don’t see it,
or if this exercise feels too much like a slog,
search the internet for the “tabular method of integration”.
-
Demonstrate how to evaluate the following integrals.
Hint: you must deftly wield some trigonometric identities.
\[
\int \sin^3(x)\cos^6(x)\,\mathrm{d}x
\qquad
\int \sin^6(x)\cos^3(x)\,\mathrm{d}x
\qquad
\int \sin^4(x)\cos^4(x)\,\mathrm{d}x
\]
\[
\int \sec^3(x)\tan^3(x)\,\mathrm{d}x
\qquad
\int \sec^4(x)\tan^4(x)\,\mathrm{d}x
\qquad
\int \sec^3(x)\tan^4(x)\,\mathrm{d}x
\]
-
Demonstrate how to evaluate the following integrals.
Hint: each one is begging you to make a trigonometric substitution.
\[
\int \frac{\sqrt{x^2+1}}{x} \,\mathrm{d}x
\qquad
\int x\sqrt{x^2+144} \;\mathrm{d}x
\qquad
\int \sqrt{x^2+12x+85} \;\mathrm{d}x
\]
\[
\int \frac{\sqrt{x^2-9}}{x^4} \,\mathrm{d}x
\qquad
\int \frac{\mathrm{d}x}{x\sqrt{7-x^2}}
\qquad
\int \sqrt{1-144x^2} \;\mathrm{d}x
\]
\[
\int\limits_{\sqrt{2}}^{2} \frac{\mathrm{d}x}{x^3\sqrt{x^2-1}}
\qquad
\int\limits_{\sqrt{0}}^{7} x^3\sqrt{x^2+49} \;\mathrm{d}x
\qquad
\int\limits_{\sqrt{0}}^{1} \sqrt{x^2+1} \;\mathrm{d}x
\]
-
Demonstrate how to evaluate the following integrals.
Hints: remember which functions
have antiderivative \(\ln\) and \(\arctan\),
remember that integrals break up across sums,
and remember the technique of completing-the-square.
\[
\int \frac{x}{x+49} \,\mathrm{d}x
\qquad
\int \frac{1}{x+49} \,\mathrm{d}x
\qquad
\int \frac{x+1}{x+49} \,\mathrm{d}x
\qquad
\int \frac{x^2+1}{x+49} \,\mathrm{d}x
\]
\[
\int \frac{x}{x^2+49} \,\mathrm{d}x
\qquad
\int \frac{1}{x^2+49} \,\mathrm{d}x
\qquad
\int \frac{x+1}{x^2+49} \,\mathrm{d}x
\qquad
\int \frac{x^2+1}{x^2+49} \,\mathrm{d}x
\]
\[
\int \frac{x}{x^2 + 12x + 85} \,\mathrm{d}x
\qquad
\int \frac{1}{x^2 + 12x + 85} \,\mathrm{d}x
\qquad
\int \frac{x+1}{x^2 + 12x + 85} \,\mathrm{d}x
\]
-
Demonstrate how to evaluate each these integrals
of a rational function.
\[
\int \frac{x - 13}{x^2 + 4x - 21} \,\mathrm{d}x
\qquad
\int \frac{5x^2 + x - 13}{(x-3)\big(x^2 + 4x - 21\big)} \,\mathrm{d}x
\]
\[
\int \frac{10x^2 + x - 13}{(x-3)\big(x^2 + 7\big)} \,\mathrm{d}x
\qquad
\int \frac{x^3 + 9x^2 - 108}{x^2 + 4x - 21} \,\mathrm{d}x
\]
-
You know that
\[ \int \frac{\mathrm{d}x}{x^2+1} = \arctan(x) + C \,.\]
But notice that \(x^2+1\) factors as
\((x+\mathbf{i})(x-\mathbf{i})\), where \(\mathbf{i}^2 = 1\).
Have no fear, \(\mathbf{i}\) is just like any other number.
Use this fact to integrate \(1/(x^2+1)\)
via its partial-fraction decomposition,
thereby obtaining a formula for \(\arctan(x)\)
in terms of logarithms of complex numbers.
-
This is a collection of integrals
that can be evaluating using some combination of
the elementary techniques discussed in our class.
Evaluate them.
\[
\int \sqrt{x}\ln(x) \,\mathrm{d}x
% by-parts
\qquad \qquad
\int x\sec(x)\tan(x) \,\mathrm{d}x
% by-parts
\qquad \qquad
\int \frac{\mathrm{e}^x}{1 + \mathrm{e}^{2x}} \,\mathrm{d}x
% substitution, arctan
\]
\[
\int \frac{\sin\left(\frac{1}{x}\right)}{x^3} \,\mathrm{d}x
%
\qquad \qquad
\int \arctan(\sqrt{x}) \,\mathrm{d}x
%
\qquad \qquad
\int \frac{\sin(x)}{1+\cos(x)} \,\mathrm{d}x
%
\]
\[
\int \frac{x^3}{x^4-1} \,\mathrm{d}x
% sub
\qquad \qquad
\int \frac{x^5}{x^4-1} \,\mathrm{d}x
% division, partial-frac
\qquad \qquad
\int \frac{x^4-1}{x^5} \,\mathrm{d}x
% elementary
\]
-
Suppose that both the ln and log buttons
have fallen off your calculator,
but you desperately need to compute
a decimal approximation of \(\ln(4)\).
-
First recall that the \(\ln\) function
is defined in terms of an integral.
Write down the definite integral that equals \(\ln(4)\).
-
Use the midpoint rule to estimate
the value of \(\ln(4)\) with six equal subintervals.
Then use the trapezoid rule to estimate \(\ln(4)\),
and then use Simpson’s rule.
How much better does each approximation get?
-
Can you write a computer program to approximate
the decimal presentation of \(\ln(4)\)?
As a reference, to check your program,
past mathematicians have computed
\[\ln(4) \approx 1.3862943611198906188344642429163531361510002687\,.\]
-
Each of the following integrals are improper.
For each integral, first identify why it’s improper.
Then, looking at the integrand, predict whether the integral
will be be convergent or divergent:
compare it to previous integrals you’ve evaluated,
and practice building your intuition on this.
Then get to work evaluating the integral if it’s convergent,
or otherwise showing it’s divergent.
Note that if it’s divergent you can save time
by comparing it to an integral you already know is divergent.
\[
\int\limits_{0}^{\pi/2} \tan(x) \,\mathrm{d}x
\qquad
\int\limits_{1}^{11} \frac{1}{x^2+34x-111} \,\mathrm{d}x
\qquad
\int\limits_{0}^{\infty} x\mathrm{e}^{-3x} \,\mathrm{d}x
\qquad
\int\limits_{2}^{3} \frac{1}{\sqrt{3-x}} \,\mathrm{d}x
\]
\[
\int\limits_{444}^{\infty} \frac{1}{(x-10000)^2} \,\mathrm{d}x
\qquad
\int\limits_{444}^{\infty} \frac{1}{(x^2-10000)^2} \,\mathrm{d}x
\qquad
\int\limits_{0}^{1} \frac{\ln(x)}{x} \,\mathrm{d}x
\qquad
\int\limits_{0}^{1} \frac{\ln(x)}{\sqrt{x}} \,\mathrm{d}x
\]
\[
\int\limits_{-\infty}^{\infty} \frac{1}{x^4}-x^4 \,\mathrm{d}x
\qquad
\int\limits_{0}^{\infty} \sin(x)\mathrm{e}^{\cos(x)} \,\mathrm{d}x
\qquad
\int\limits_{0}^{1} \frac{\mathrm{e}^{1/x}}{x^4} \,\mathrm{d}x
\qquad
\int\limits_{0}^{\infty} \frac{\mathrm{e}^x}{\mathrm{e}^{2x}+3} \,\mathrm{d}x
\]
-
For \(R \gt 1\), define the function
\[A(R) = \int\limits_1^R \frac{1}{x^{p}} \,\mathrm{d}x\]
-
Show that if \(p=1\), then \(A(R) = \ln|z|\),
but for \(p\neq 1\), we instead have
\[ A(R) = \frac{1}{1-p}\left(R^{1-p}-1\right)\,. \]
-
Show that if \(0 \lt p \leq 1\),
then \(\lim\limits_{R \to \infty} A(R) = \infty\).
-
Show that if \(p \gt 1\),
then \(\lim\limits_{R \to \infty} A(R) = \frac{1}{p-1}\).
Look out for the vocabulary term \(p\)-series
later in the course, and remember this exercise when you see it.
-
Draw a picture of the region between the curves \(y = \ln(x)\) and
\(y = \ln\left(x^2\right)\) in the lower half-plane.
Is the area of this region finite?
If it is, what’s its area? If not how do you know?
Write down an integral that represents the volume of the solid
formed by rotating this region about the \(y\)-axis.
Is the volume of this solid finite?
Challenges
-
Spivak (mostly)
Demonstrate how to evaluate the following integrals.
\[
\int \frac{\arctan(x)}{1+x^2} \,\mathrm{d}x
\qquad
\int \frac{x\arctan(x)}{(1+x^2)^3} \,\mathrm{d}x
\qquad
\int \ln\big(\sqrt{1+x^2}\big) \,\mathrm{d}x
\]
\[
\int x\ln\big(\sqrt{1+x^2}\big) \,\mathrm{d}x
\qquad
\int \frac{x^2-1}{x^2+1}\cdot\frac{1}{\sqrt{1+x^4}} \,\mathrm{d}x
\qquad
\int \arcsin\!\big(\sqrt{x}\big) \,\mathrm{d}x
\]
\[
\int \frac{x}{1+\sin(x)} \,\mathrm{d}x
\qquad
\int \mathrm{e}^{\sin{x}} \cdot \frac{x\cos^3(x)-\sin(x)}{\cos^2(x)} \,\mathrm{d}x
\qquad
\int \frac{\mathrm{d}\theta}{1-\sin^4(\theta)}
\]
\[
\int \sqrt{\tan(x)} \,\mathrm{d}x
\qquad
\int \frac{\mathrm{d}x}{\sqrt{1+\mathrm{e}^{2x}}}
\qquad
\int \ln\left(\mu + \sqrt{1-\mu}\right)\,\mathrm{d}\mu
\]
-
For which values of \(p\) do these integrals converge?
For those values of \(p\) what do these integrals converge to?
\[
\int\limits_{0}^{1} \frac{\ln(x)}{x^p} \,\mathrm{d}x
\qquad \qquad
\int\limits_{\mathrm{e}}^{\infty} \frac{1}{x\big(\ln(x)\big)^p} \,\mathrm{d}x
\]
-
Find the interval \([a,b]\)
for which the value of this integral is maximal:
\[\int\limits_a^b (2 + x - x^2) \,\mathrm{d}x\]
-
If \(f\) is a differentiable function such that
\(\int_0^x f(t) \,\mathrm{d}t = \big(f(x)\big)^2\) for all \(x\),
find a formula for \(f\).
-
Consider a square of side-length two,
and consider the region consisting of all points in the square
that are closer to the center of the square
than to any side of the square.
What is the area of this region?
-
Evaluate \(\int \frac{\mathrm{d}x}{x^7-x}\).
The straightforward approach would be to start
with partial fractions, but that would be brutal.
Try a substitution.
-
Show that the integral
\[ \int\limits_{0}^{\infty} \frac{\ln(x)}{1+x^2} \,\mathrm{d}x \]
evaluates to zero by breaking up its domain of integration into two parts
based on where the integrand is negative or positive,
and using the substitution \(x \leftrightarrow \frac{1}{x}\)
on one of those parts.
Then use this same trick to show that the value of the following
integral doesn’t depend at all on the value of the real number \(p\).
\[ \int\limits_{0}^{\pi/2} \frac{\mathrm{d}\theta}{1+\big(\tan(\theta)\big)^p} \]
-
Evaluate
\(\int_0^1 \sqrt[3]{1-x^7} - \sqrt[7]{1-x^3} \,\mathrm{d}x\,.\)
-
If \(n\) is a positive integer, prove that
\[\int\limits_0^1 \big(\ln(x)\big)^n \,\mathrm{d}x = (-1)^n n!\,.\]
-
Putnam
Evaluate these integrals.
\[
\int\limits_0^1 \frac{\ln(x+1)}{1+x^2} \,\mathrm{d}x
\qquad
\int\limits_2^4 \frac{\sqrt{\ln(9-x)}}{\sqrt{\ln(9-x)}+\sqrt{\ln(x+3)}} \,\mathrm{d}x
\]