MTH 122 — Calculus II
Essex County College — Division of Mathematics and Physics
1
Lecture Notes #7 — Sakai Web Project Material
1
Trigonometric Substitutions,
§
7.3
In this section of the textbook we will be presented with difficult integrands that contain a
function of
x
that looks like:
p
a
2

x
2
,
p
a
2
+
x
2
,
p
x
2

a
2
.
In general we will be looking to make a substitution, but in these cases our substitution will be
another function with another parameter. To make our calculations simpler, we need to make
these
new
functions invertible, that is, they should be
onetoone
. For example, here we’re using
a new invertible function
g
(
θ
) to rewrite the integrand
f
(
x
):
Z
f
(
x
) d
x
=
Z
f
(
g
(
θ
))
g
0
(
θ
) d
θ,
which really looks like classic
u
substitution, and it may even be referred to as
inverse
u

substitution
by some in the
mathematical digerati
. So let’s take a look at some examples, and I
will divide it into three cases.
1. The integrand contains an expression of the form
√
a
2

x
2
, just use
x
=
a
sin
θ
. Now as
you already know, the sine function is not invertible, so we need to restrict the domain of
θ
∈
[

π/
2
, π/
2], and will use 1

sin
2
θ
= cos
2
θ
to help simplify the integral. It should
be noted that in general,
√
cos
2
θ
=

cos
θ

, but for the restricted domain of theta, we have
√
cos
2
θ
= cos
θ
.
Example:
Z
p
1

9
x
2
d
x
= 3
Z
p
1
/
9

x
2
d
x
Now let
x
=
sin
θ
3
, then 3 d
x
= cos
θ
d
θ
.
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 Spring '10
 Ban
 Calculus, Trigonometry, Division, Mathematica, Inverse function, dθ

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