Math 222 Calculus II
Integration: When to use what
If you’re presented with an integral that you don’t know how to solve, there is no perfect system
to figure it out. However, there are some standard things to try.
1. Always check for a u-substitution first! When a u-substitution works, it generally makes
things much easier.
Z
4x − 2
Example:
dx can be done by partial fractions, but with the substitution u = x2 −x,
x2 − x Z
2
the integral becomes
du, which is easy to evaluate.
u
2. See if there is a way to simplify the integrand
Z
Z
1 + x4
1
Example:
dx =
+ x3 dx
x
x
3. Try trigonometric identities for integrals involving trigonometric functions.
Example: Using the identity 1 + tan2 θ = sec2 θ, we can write
Z
Z
tan4 θ sec4 θ dθ as
tan4 θ(1 + tan2 θ) sec2 θ dθ, which can be evaluated by letting u = tan θ.
√
√
√
4. Try trigonometric substitutions for integrals involving x2 + a2 , x2 − a2 , or a2 − x2 . Trig
substitutions sometimes also work for integrals involving x2 + a2 , x2 − a2 , or a2 − x2 .
5. Try integration by parts for integrals of products.
6. Try partial fractions for integrals of rational functions
polynomial polynomial
.
7. If nothing seems to work, see what a computer algebra system (such as Maple, Mathematica,
or Derive) returns as an answer. Wolfram Alpha (www.wolframalpha.com) is free and can do
integration.
There are some dangers with relying on computer algebra systems. One is that they may
return answers in aZ form that is not easy to work with. For example, if you ask Wolfram
p
Alpha to integrate
1 + x2 dx, you will get the following result:
Z p
p
1 + x2 dx = 21 (x 1 + x2 + sinh−1 (x)) + C
Here, sinh−1 (x) is the inverse hyperbolic sine function, which is not a function you are probably used to dealing with. In fact, a nicer form for the answer is
Z p
p
1 + x2 dx = 21 (x 1 + x2 + ln(x +
p
1 + x2 )) + C
A second danger
is that computer
algebra systems have limitations. For example, when asked
Z
q
to integrate
(1 + ln(x)) 1 + (x ln(x))2 dx, Wolfram Alpha says: “no result found in terms
of standard mathematical
functions”. However, via the substitution u = x ln(x), the above
Z
integral becomes
p
1 + u2 du, an integral that can be handled using the integration formula
from the previous paragraph.