MA 16021
Strategies for choosing an integration technique
K. Rotz
On an exam, you normally won’t be told what technique to use to evaluate an integral. It’s up to
you to figure that out. We talked about a few techniques in class – u-sub, integration by parts, and
partial fractions – so here are some guidelines to figure out what to use.
1. List of basic integrals. You can only perform an integration step if the integral looks like one
of the following that we covered in class.
Z
Z
1
xn+1
n
+ C (if n 6= −1)
dx = ln |x| + C
x dx =
n+1
x
Z
Z
1
x
1
ex dx = ex + C
dx = tan−1 ( ) + C
2
2
x +a
a
a
Z
Z
sin x dx = − cos x + C
cos x dx = sin x + C
Z
Z
sec2 x dx = tan x + C
csc x cot x dx = − csc x + C
Z
Z
sec x tan x dx = sec x + C
csc2 x dx = − cot x + C
The rest of the techniques that we covered (u-sub, integration by parts, and partial fractions) are
all ways of boiling down a given integral into one of these forms. If you’re trying to integrate
something that doesn’t look like one of these, you’ve got some work to do!
Memorizing these is definitely recommended, as the exam will be next to impossible without them.
2. “Integral with a twist”: If you have an integral that almost looks like one of the above, just
with a linear mx + b inside of the function, you can do the integration as usual and just divide
the final result by the slope of the line, m. I like to call this “integration with a twist,” the twist
being the dividing by m part. For example,
Z
1
e2x dx = e2x + C.
2
You integrate the exponential as normal (giving e2x , since the integral of the exponential function
is itself), but then divide by the coefficient of the x-term which appears inside the exponential, 2,
to correct your guess. The 21 is called the “twist factor.” Other examples:
twist
Z
sine integral
z}|{
}|
{
1 z
sin(4x + 3) dx = ( ) (− cos(4x + 3)) +C
4
Z
1 (5x − 9)11
(5x − 9)10 dx = ( )
+C.
5
11
|{z} | {z }
twist
1
power rule
MA 16021
Strategies for choosing an integration technique
K. Rotz
Note that in each case, the b part of the mx + b does not matter to the twist factor. Only the
slope m matters!
The basis for integrals with a twist is u-substitution: if you let u = mx + b, then du = m dx, or
dx = m1 du. So when you do the substitution step, you kick out a factor of m1 , the twist factor.
Be careful not to get carried away with this! This only works for linear functions. It doesn’t
extend to quadratics or higher order polynomials. In other words, things like the following just do
no hold.
Z
cos(x2 )
2
+C
sin(x ) dx 6= −
2x
3. Simplify if possible. Sometimes you can distribute through powers of x into parentheses, or
cancel out stuff from fractions. For example,
Z
Z
√
x( x − 1) dx = (x3/2 − x) dx
Z
Z
Z 2
x(x + 1)
x +x
dx =
dx = x dx.
x+1
x+1
4. Integrals with trig in them. If you have to integrate something with just trigonometric functions
in it, ask yourself the following: do I know some function whose derivative is equal to that function?
For example, if I need to integrate csc2 x, I know that the derivative of cot x is − csc2 x, and
therefore the integral of csc2 x is − cot x. If the answer to that question is no, then it’s a good idea
to write the integral in terms of sines and cosines and look for a u-substitution. For example,
Z
tan x
dx
cos3 x
doesn’t have an obvious anti-derivative or even an obvious candidate for u-sub. But, if you write
sin x
tan x = cos
, it becomes
x
Z
sin x
dx
cos4 x
and this has the obvious u-sub u = cos x.
5. Look for an “obvious” u-substitution. This typically involves looking at the denominator of
fractions, or looking inside a function or set of parentheses. If either of those has a derivative which
is also a multiplicative factor of the integral (apart from perhaps a constant factor), u-substitution
is a great choice.
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MA 16021
Strategies for choosing an integration technique
K. Rotz
Examples:
Z
2
xe−x dx
Z
sin x sin(cos x) dx
Z
x2
3x + 3
dx.
+ 2x − 1
In the first, u = −x2 is inside of another function (the exponential), and its derivative is an x-term
(which also appears as a multiplicative factor of the integral). So u = −x2 is a good substitution
here.
In the second, u = cos x appears inside another function, and its derivative, u0 = − sin x, is, apart
from a negative, also in the integral as a multiplicative factor.
For the third, u = x2 + 2x − 1 is in the denominator. Its derivative is u0 = 2x + 2 = 2(x + 1). That
doesn’t immediately look
it appears in the numerator of the integral, but if you factor out a
R like
du
x+1
dx. So in this case, once you solve for dx = 2(x+1)
and substitute
3 first, the integral is 3 x2 +2x−1
3
this and u in, there will be a factor of 2 in front of the integral.
Here is an example of where u-sub just doesn’t work
Z
2
x3 ex dx,
u = x2 is the “obvious” first choice for a u-sub since it is inside of the exponential function.
However, its derivative is du = 2x dx, and that’s not a multiplicative factor of the integral. You
might object “But x3 = x2 · x, so there is a factor of x in the integral!” That’s true, but x3 is not
a constant multiple of x.
You could also protest by saying, “But hey, if I pick u = x3 , then its derivative is an x2 term, and
that appears in the integral.” And again, you’d be right, that is in the integral. But it’s not a
multiplicative factor of the integrand – indeed, the x2 is inside of the exponential – and that’s bad
for u-sub. It’s absolutely crucial that the u0 be a multiplicative factor of the integrand.
6. Partial fractions. If u-sub fails and the integrand is a fraction involving polynomials (a.k.a. a
rational function), go into partial fraction mode. If the degree of the numerator is greater than
or equal to the degree of the denominator, do polynomial long division first and then perform
partial fractions on the remainder term. Lots of times these sorts of problems have logarithms and
1
arctangents in their final answers from integrating terms like x+1
and x25+4 , respectively.
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MA 16021
Strategies for choosing an integration technique
K. Rotz
7. Integrate by parts. If u-sub fails and it’s not a rational function, resort to integration by parts
as a last resort. The first thing that should pop in to your head with integration by parts is the
acronym LATE to help you choose your u and dv.
There are two cases where you actually want to skip straight to integration by parts
Z
• (positive integer power of x)×(something you can integrate repeatedly) dx. Here, the term
“something you can integrate repeatedly” typically means an exponential, sine or cosine, e.g.
e−5x , sin 10x, or cos 3x. In each of those cases, you can integrate with a twist (see tip 2)
over and over again. But be careful! You need to have only an x inside of the term you’re
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intending to integrate repeatedly. For example, you can’t integrate cos(x2 ) or ex at all, let
alone multiple times.
Examples of this type:
Z
Z
3 −5x
xe
dx,
x sin x dx
both fit the bill since you can integrate e−5x and sin x repeatedly.
Z
• (a power of x) × (a logarithm) dx. Here, the power of x could be positive, negative, or even
fractional. So if you see something like this
Z
√
x ln(x4 ) dx,
which is a modification of the problem from quiz 3, or even something as hideous looking as
Z
(x2 + 3x + 5) ln(2x3 + 1) dx,
each of those are just asking to be integrated by parts. One exception to this exception is if
the power of x is −1, in other words, if you’re integrating
Z
ln x
dx.
x
In that case, a quick u = ln x substitution will give you the answer.
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