2Integration
IA Differential Equations
2.2 Methods of integration
Integration is substantially harder than differentiation. Given an expression,
the product rule and chain rule combined with a few standard derivatives can
usually differentiate it with ease. However, many seemingly-innocent expressions
can turn out to be surprisingly difficult, if not impossible, to integrate. Hence
we have to come up with a lot of tricks to do integrals.
Example (Integration by substitution). Consider
R
1−2x
√
x−x
2
d
x
. Write
u
=
x−x
2
and du = (1 − 2x) dx. Then the integral becomes
Z
du
√
u
= 2
√
u + C = 2
p
x − x
2
+ C.
Trigonometric substitution can be performed with reference to the following
table: if the function in the 2nd column is found in the integrand, perform
the substitution in the 3rd column and simplify using the identity in the first
column:
Useful identity Part of integrand Substitution
cos
2
θ + sin
2
θ = 1
√
1 − x
2
x = sin θ
1 + tan
2
θ = sec
2
θ 1 + x
2
x = tan θ
cosh
2
u − sinh
2
u = 1
√
x
2
− 1 x = cosh u
cosh
2
u − sinh
2
u = 1
√
1 + x
2
x = sinh u
1 − tanh
2
u = sech
2
u 1 −x
2
x = tanh u
Example. Consider
R
√
2x − x
2
d
x
=
R
p
1 − (x − 1)
2
d
x
. Let
x −
1 =
sin θ
and thus dx = cos θ dθ. The expression becomes
Z
cos
2
θ dθ =
Z
cos 2θ + 1
2
dθ
=
1
4
sin 2θ +
1
2
θ + C
=
1
2
sin
−1
(x − 1) +
1
2
(x − 1)
p
2x − x
2
+ C.
Theorem (Integration by parts).
Z
uv
0
dx = uv −
Z
vu
0
dx.
Proof.
From the product rule, we have (
uv
)
0
=
uv
0
+
u
0
v
. Integrating the whole
expression and rearranging gives the formula above.
Example. Consider
R
∞
0
xe
−x
d
x
. Let
u
=
x
ad
v
0
=
e
−x
. Then
u
0
= 1 and
v = −e
−x
. We have
Z
∞
0
xe
−x
dx = [−xe
−x
]
∞
0
+
Z
∞
0
e
−x
dx
= 0 + [−e
−x
]
∞
0
= 1
Example (Integration by Parts). Consider
R
log x
d
x
. Let
u
=
log x
and
v
0
= 1.
Then u
0
=
1
x
and v = x. So we have
Z
log x dx = x log x −
Z
dx
= x log x − x + C