TECHNIQUES OF INTEGRATION
MAX GOERING – MARCH 17, 2017
Abstract. Hello! Welcome to Math 307. If done correctly, Differential Equations can be one of the
most rewarding math courses you’ve taken. Differential equations are used throughout all disciplines
are natural and social sciences, and therefore the applications are varied and should interest everyone.
However, integration is critical to successfully learning differential equations. It would be a shame if
your interest in differential equations were stifled by a weak background in integration. These notes are
to emphasize the importance of techniques of integration. Of most importance is probably Integration
by Partial Fractions. I thought this was important enough to just include outside references that are
more detailed that the notes I have on other techniques of integration.
1. Techniques of Integration
Reading mathematics is a skill, that once practiced will save you lots of time. Textbooks are
written very carefully to include a lot of information in very little space. This means, when reading
a textbook, one typically needs scratchpaper, and to work alongside the book as they read.
Nonetheless, reading a text book is a skill. If you haven’t practiced it, it is daunting. I STILL
SUGGEST YOU PRACTICE. However, if you are very rusty on integration, it may be a nice warmup to watch videos posted at patrickJMT.com before reading this note.
In an introduction to differential equations course you will learn a myriad of techniques used to
solve simple differential equations. The you learn so many different techniques is because a small
change in a differential equation, can have great impact on the method used to solve it. However,
the most reoccurring theme in solving differential equations is the necessity to compute integrals.
Hence, we review some techniques of integration.
1.1. Integration by partial fractions. Is it rude to immediately just reference you somewhere
else? I hope not. For videos discussing integration by partial fraction decomposition check out the 8
videos by patrickJMT at patrickJMT.com. Alternatively, here are some other places to look
1) Most obviously, look at your old calculus book and then at your calculus notes if you have
lingering questions.
2) If you know how to do partial fractions but just want to speed up, then consider learning the
techniques found here:
• The cover-up method for linear factors
• The cover-up method with irreducible quadratics
• Don’t forget about comparing coefficients if you have a lot of ”repeated” factors.
Tried these things and still no luck? Email me to set-up an appointment or swing by my office
hours.
Date: March 17, 2017.
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MAX GOERING – MARCH 17, 2017
1.2. Integration by Parts. One of the most fundamental tools used to evaluate integrals in Integration by parts. We usually recall integration by parts as the formula:
Z
Z
(1.1)
u dv = uv − v du.
While this is typically enough to remember, let’s not forget that we require the functions u and v
to satisfy some properties. Namely, u, u0 , v, v 0 need to be continuous. A more precise statement of
integration by parts would be
Z b
Z b
x=b
0
v(x)u0 (x) dx.
(1.2)
u(x)v (x) dx = u(x)v(x) x=a −
a
a
R1
x
Example 1.1.
R Consider the integral 0 xe dx. We know how to differentiate x, and we know how
to integrate et dt, so we can consider doing integration by parts with
(
u(x) = x u0 (x) = dx
v(x) = ex v 0 (x) = ex
Then, making these substitutions in Equation (1.2) we learn
Z 1
Z 1
x
x x=1
xe dx = xe x=0 −
ex dx = (e1 − 0) − (e1 − e0 ) = 1
0
0
Another,
1.3. Trigonometric Integrals. Before we discuss trigonometric integrals, we should recall some
important trigonometric identities. Perhaps the most important trigonometric identity is
(1.3)
cos2 (θ) + sin2 (θ) = 1.
From here we can divide both sides by cos2 (θ) to learn another important identity
(1.4)
1 + tan2 (θ) = sec2 (θ).
The point of this section, is to learn how to efficiently evaluate integrals of the form
Z
Z
m
n
sin (x) cos (x) dx
and
tanm (x) secn (x) dx,
for integers m and n.
R
1.3.1. Computing sinm (x) cosn (x) dx.
The key to evaluating this integral is to use Equation (1.3) in combination with the basic derivatives
d d sin(x) = cos(x)
and
cos(x) = − sin(x).
dx
dx
Case 1. If m is odd.
In this case, we write m = 2k + 1, and use Equation (1.3) to write
k
k
sinm (x) = sin2k (x) sin(x) = sin2 (x) sin(x) = 1 − cos2 (x) sin(x),
then making the substitution u = cos(x) we know du = − sin(x) so that
Z
Z
m
n
sin (x) cos (x) dx = (1 − cos2 (x))k cosn (x) sin(x) dx
Z
= − (1 − u2 )k un du,
TECHNIQUES OF INTEGRATION
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then one can expand out (1 − u2 )k and evaluate the integral using the power rule.
Case 2. If n is odd. In this case, we write n = 2` + 1, and use Equation (1.3) to write
cosn (x) = (1 − sin2 )` cos(x).
This time, use the substitution v = sin(x) and dv = cos(x) to discover
Z
Z
m
n
sin (x) cos (x) dx = (1 − sin2 (x))` sinm (x) cos(x) dx
Z
= (1 − v 2 )` v m dv,
again, one can sovle this integral by expanding out the (1 − v 2 )` term.
Case 3. Both m and n are even. In this case, there are two trains of thought on how to proceed.
Technically, they work better depending on how large m and n are. The more common technique, is
to write m = 2k and use Equation (1.3) to write
sinm (x) = (sin2 (x))m = (1 − cos2 (x))m ,
expand out and multiply by cosn (x), then reduce all of your even powers of cos(x) to cos(x) to the
power of 1 by repeatedly using the identity:
1
(1.5)
cos2 (x) = (1 + cos(2x)).
2
This method works great if (m, n) = (0, 2) or (m, n) = (2, 0). However, it’s hardly tractible if, for
instance m = 4 and n = 2.
A less common but in my opinion substantially nicer technique is to use the the ”Sine
and Cosine reduction formulas”. These formulas are can be shown to be true by applying integration
by parts. In fact, they do not even require m, n to be even integers. The formulas are as follow (if
you want to know where they come from and cannot re-create them yourselves, feel free to first check
the index of your calculus text book for ”reduction formula(s)” and if you cannot figure it out after
reading the book, don’t hesitate to email me).
(R
R
sinn (x) dx = − n1 cos(x) sinn−1 (x) + n−1
sinn−2 (x) dx
n
R
R
cosn (x) dx = n1 cosn−1 (x) + n−1
cosn−2 (x) dx
n
where n ≥ 2 is an integer.
R
Example 1.2. Compute the indefinite integral sin4 (x) cos2 (x) dx.
We first apply the identity in Equation (1.3) to re-write cos2 (x) = 1 − sin2 (x). Then, we have
Z
Z
Z
4
2
4
(1.6)
sin (x) cos (x) dx = sin (x) dx − sin6 (x) dx.
Since the reduction
formula reduces the power of sin inside our integral by 2, we first check out
R
the integral sin6 (x) dx and we will be left with something that we can combine with our integral
sin4 (x) dx. That is, applying the reduction formula for sine we learn
Z
Z
1
5
6
5
sin (x) dx = − cos(x) sin (x) +
sin4 (x) dx
6
6
so
Z
Z
Z
1
5
4
6
5
(1.7)
sin (x) dx − sin (x) dx = cos(x) sin (x) + (1 − ) sin4 (x) dx.
6
6
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MAX GOERING – MARCH 17, 2017
Now, we apply the reduction formula on the remaining integral to learn
Z
Z
3
1
3
4
sin2 (x) dx
sin (x) dx = − cos(x) sin (x) +
4
4
and one last application of the reduction formula tells us that up to a constant +C, we have
Z
Z
1
1
1
1
2
1
sin (x) dx = − cos(x) sin (x) +
1 dx = − cos(x) sin(x) + x
2
2
2
2
So, combining the previous two equations yields:
3
1
1
1
3
4
− cos(x) sin(x) + x .
sin (x) dx = − cos(x) sin (x) +
4
4
2
2
Finally, combining this with Equations (1.6) and (1.7) yields
Z
Z
Z
4
2
4
sin (x) cos (x) dx = sin (x) dx − sin6 (x) dx
Z
1
5
5
= cos(x) sin (x) + (1 − ) sin4 (x) dx
6
6
1
1
= cos(x) sin5 (x) +
6
6
!
1
3
1
1
3
− cos(x) sin (x) +
− cos(x) sin(x) + x
+ C.
4
4
2
2
Since these notes are not algebraically motivated, we leave reducing this expression as an exercise to
the interested reader.
1.3.2. Thank you for still reading. Thank you for taking the time to read the notes I wrote for you all.
Your reward for reading this far is that you, and the few of your dedicated classmates who also read
this much know that there will be a pop quiz on Wednesday over techniques of integration. Waithow is it a pop-quiz if I’m telling you it exists? Exactly. It’s not. For you. Potentially only you
know that the quiz on Wednesday will include a simple integration by parts and a simple integration
by partial fractions problem.
R
1.3.3. Computing tanm (x) secn (x) dx. Here, the key idea is to combine Equation (1.4) with the
basic derivative rules:
d d tan(x) = sec2 (x)
and
sec(x) = sec(x) tan(x).
dx
dx
Case 1. If m is odd and n ≥ 1. In this case, we write m = 2k + 1 and use Equation (1.4) to write
tanm (x) = tan2k (x) tan(x) = (sec2 (x) − 1)k tan(x)
and the u substitution u = sec(x) so that du = sec(x) tan(x) dx. Then,
Z
Z
m
n
tan (x) sec (x) dx = (sec2 (x) − 1)k secn−1 (x) sec(x) tan(x) dx
Z
= (u2 − 1)k un−1 du,
again from here one simply expands out the (u2 − 1)k term then integrates.
Case 2. If n, m are both even and n ≥ 2.
In this case, we simply write n = (n − 2) + 2 = 2` + 2. Then, using Equation (??) we write
secn (x) = (sec2 (x))` sec2 (x) = (1 + tan2 (x))` sec2 (x).
TECHNIQUES OF INTEGRATION
Then, using the substitution v = tan(x) so that dv = sec2 (x) dx we attain
Z
Z
m
n
tan (x) sec (x) dx = (1 + tan2 (x))` tanm (x) sec2 (x) dx
Z
= (1 + v 2 )` v m dv,
again, we expand (1 + v 2 )` then integrate.
Case 3. If neither of the above apply, change to sines and cosines and hope it works!
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