Integration Strategies
While it can be relatively systematic to differentiate arbitrary expressions, it is typically much more difficult
to integrate. In this section we give intuition and strategies for evaluating integrals.
First and foremost is knowledge of the basic antiderivative formulas. The common antiderivative formulas
in the following table should be committed to memory.
Common Antiderivatives
Z
1
1
n
n+1
• x dx =
x
+ C, n 6= −1
•
dx = ln |x| + C
n+1
x
Z
Z
• sin(x) dx = − cos(x) + C
• cos(x) dx = sin(x) + C
Z
Z
•
Z
2
sec (x) dx = tan(x) + C
•
sec(x) tan(x) dx = sec(x) + C
•
Z
•
Z
Z
•
Z
•
Z
•
csc2 (x) dx = − cot(x) + C
csc(x) cot(x) dx = − csc(x) + C
Z
sec(x) dx = ln | sec(x) + tan(x)| + C
•
1
dx = arctan(x) + C
x2 + 1
•
ex dx = ex + C
•
√
1
dx = arcsin(x) + C
1 − x2
Z
x2
Z
x
1
1
dx = arctan
+C
2
+a
a
a
ax dx =
ax
+C
ln(a)
If you do not recognize the antiderivative of a function, the following steps walk you through the thought
process involved with integration. For each strategy several ‘partial’ examples will be used to illustrate the
process. A large collection of practice problems can be found at the end of this section.
Integration Strategies:
1. Is it a standard integral? Or a modified version of a standard integral?
Example:
Z
Z
5
1
•
dx = 5
dx = 5 ln |x| + C.
x
x
Z
Z
Z
1
1
1
1
√
p
p
•
dx =
dx =
dx and the resulting integral
2
2
2
2
a
a −x
a (1 − (x/a) )
1 − (x/a)2
x
can be evaluated via a combination of the substitution u = a and the common antiderivative
Z
1
√
dx = arcsin(x) + C.
1 − x2
2. Algebraically simplify the integrand.
Example:
1
Calculus II Resources
Integration Techniques
Z
Z
(1 + ex )2
1 + 2ex + e2x
dx
=
dx
=
e−x + 2 + ex dx
ex
ex
Z
Z
Z
1
x−1+1
x
dx =
dx = 1 +
dx
•
x−1
x−1
x−1
Z
•
3. Look for obvious substitutions.
Example:
Z
Z
1
1
arctan(x)
•
dx
=
arctan(x) 2
dx Use u = arctan(x) so that du = 2
dx.
x2 + 1
x +1
x +1
Z 1/x
1
e
dx Use u = x1 so that du = − 2 dx.
•
x2
x
4. Classify the integrand according to its form.
• For trigonometric functions see Section ??.
• For rational functions see Section ??.
• For functions of the form x2 cos(x), x2 ex , ex sin(x), arcsin(x), ln(x), and other inverse functions
apply IBPs. See Section ??.
• For functions containing the expressions x2 − a2 , x2 + a2 , or a2 − x2 try a Trig Substitution. See
section ??.
5. Look for more subtle substitutions.
Example
√
Z
e
x
Z
eu 2u du
Z
= 2 ueu du
dx =
x = u2 , dx = 2udu
and we can now apply IBPS.
6. Try IBPs
7. Manipulate the Integrand.
Example:
Z
ex
dx =
x
e + e−x
x
ex
e
dx
ex + e−x ex
Z
ex ex
=
dx
e2x + 1
Z
and we can now use the substitution u = ex .
8. Use multiple methods
Z 1
√
Example 1. Evaluate
(1 + x)8 dx.
0
Intuition: No algebraic simplification or substitution is obvious and the integrand does not fit any of the
standard forms. The next
√ strategy is to look for a more subtle substitution. In order to pick u as an ‘inside’
function we set u = 1 + x.
2
Calculus II Resources
Integration Techniques
√
Let u = 1 +
√
x=u−1
x so that x = (u − 1)2
dx = 2(u − 1)du
Z
1
(1 +
√
x = 0√
u=1+ 0
u=1
x)8 dx = 2
2
Z
0
x = 1√
u=1+ 1
u=2
u8 (u − 1) du
1
2
Z
u9 − u8 du
=2
1
2
1 10 1 9 =2
u − u 10
9
10
1
2
29
1
1
=2
−
−
−
10
9
10 9
Observation: This is a difficult integral! You may have to attempt multiple methods before finding a
successful approach. This is a large part of the learning process.
See solution video
Z
Example 2. Evaluate
sin(2x) ln(sin(x)) dx.
Intuition: There does not appear to be any ‘obvious’ algebraic simplification. The next strategy is to look
for any obvious substitutions. In this example, the obvious substitution is the ‘inside’ function sin(x).
Let w = sin(x) so that dw = cos(x)dx. To rewrite the integrand in terms of the variable w we use the
double-angle formula sin(2x) = 2 sin(x) cos(x).
Z
Z
sin(2x) ln(sin(x)) dx =
2 sin(x) cos(x) ln(sin(x)) dx
Z
= 2 w ln(w) dw
The resulting integral is well-suited for IBPs. Make sure that your choice for dv is a something that you can
antidifferentiate.
u = ln(w) −→ v = 12 w2
we have
Using
du = w1 dw
.
dv = wdw
Z
Z
sin(2x) ln(sin(x)) dx = 2
w ln(w) dw
Z
= w2 ln(w) − w dw
1
= w2 ln(w) − w2 + C
2
1
= sin2 (x) ln(sin(x)) − sin2 (x) + C
2
Observation: Many examples require multiple techniques of integration.
See solution video
Summary:
If you do not recognize the antiderivative of a function, the following steps walk you through the thought
process involved with integration.
Integration Strategies:
3
Calculus II Resources
Integration Techniques
1. Is it a standard integral? Or a modified version of a standard integral?
2. Algebraically simplify the integrand.
3. Look for obvious substitutions.
4. Classify the integrand according to its form.
5. Look for more subtle substitutions.
6. Try IBPs
7. Manipulate the Integrand.
8. Use multiple methods
Practice Problems: Evaluate the following Integrals
Z
(1)
Z
(2)
Z
(3)
Z
(4)
(6)
0
Z
(9)
1
dx
2
(x + 4)2
(10)
√
1
(8)
3
x
Z
Z
x5 e−x dx
3
tan(x) sec3 (x) dx
(7)
arctan(x)
dx
x2
Z
(5)
Z
x
√
dx
1 − x4
Z
Z
x − 1 dx
(11)
x4 − 5x3 + 6x2 + 18
dx
x3 − 3x2
Z
(12)
See Solutions
4
x
dx
ex
ex cos(x) dx
√
1
ex
−1
dx
x5x dx
√
4x2
1
dx
− 4x − 3