Integrals of Even or Odd Functions with applications to Trig
Secant Substitution
1. If f (x) is an odd function, then
Z
f (x) dx = F (x) + C
for some even function F (x). If f (x) is an even function, then
Z
f (x) dx = G(x) + C
for some odd function G(x).
Proof.
We first consider the case that f is defined on (−∞, ∞). Suppose
f (x) is odd. Let H(x) be an antiderivative of f . Then H(−x) is also
an antiderivative of f since (H(−x))0 = −H 0 (−x) = −f (−x) = f (x).
Thus H(−x) = H(x) + C1 for some constant C1 . Putting x = 0, we
find that C1 = 0. So H is even.
Next assume that f (x) is even. Let H(x) be an antiderivative of
f . Then −H(−x) is also an antiderivative of f since (−H(−x))0 =
H 0 (−x) = f (−x) = f (x). Thus −H(−x) = H(x) + C1 for some
constant C1 . Putting x = 0, we find that C1 = −2H(0). It follows
that K(x) = H(x) − H(0) is an odd antiderivative of f .
Consider now the case when the domain of f is of the type I ∪ (−I)
where I = (a, ∞) for some a ≥ 0 or I = [a, ∞) for some a > 0. We
consider only the later case below, the other case being similar.
Suppose f (x) is odd, defined on (−∞, −a] ∪ [a, ∞) where a ≥ 0. Let
H(x) be an antiderivative of f on [a, ∞). Extend H(x) to (−∞, −a]
by defining H(x) = H(|x|) = H(−x) for x ∈ (−∞, −a]. Then H is
even, and for x ∈ (−∞, −a], we have
H 0 (x) = (H(−x))0 = −H 0 (−x) = −f (−x) = f (x).
This proves part one.
Suppose f (x) is even, defined on (−∞, −a] ∪ [a, ∞) where a ≥ 0. Let
H(x) be an antiderivative of f on [a, ∞). Extend H(x) to (−∞, −a]
by defining H(x) = −H(−x) for x ∈ (−∞, −a]. Then H is odd, and
for x ∈ (−∞, −a], we have
H 0 (x) = −(H(−x))0 = H 0 (−x) = f (−x) = f (x).
This proves part two.
1
R
2. When the integrand f (x) in f (x) dx involves x2 − a2 , sometimes
we√make trig substitution x = a sec θ,√
0 ≤ θ ≤ π, θ 6= π/2. For x ≥
a, x2 − a2 = a tan θ and for x ≤ −a, x2 − a2 = a| tan θ| = − tan θ.
So many times, we have to consider each case separately. But most of
the time, the integrand is either even or odd, so we can use the above
theorem, consider only the case x ≥ a, and then extend the result
to an odd or even function. This is illustrated in the following two
examples.
3. Evaluate
Z
x2
√
1
dx
x2 − a2
where a > 0 is a constant.
We consider the integral √
on [a, ∞). Substituting x = a sec θ, we have
dx = a sec θ tan θdθ, and x2 − a2 = a tan θ. So the integral becomes
√
Z
1
1 x2 − a2
1
cos θ dθ = 2 sin θ = 2
+ C.
a2
a
a
x
Since the integrand is even, and the odd extension of
the same form, we obtain
√
Z
1
1 x2 − a2
√
dx = 2
+C
a
x
x2 x2 − a2
1
a2
√
x2 −a2
x
has
valid for x positive or negative.
4. Evaluate
Z
1
dx
x(x2 − a2 )3/2
where a > 0 is a constant.
This time the integrand is odd. We consider the integral on [a, ∞).
Substituting x = a sec θ, we have dx = a sec θ tan θdθ, and (x2 −
a2 )3/2 = a3 tan3 θ. So the integral becomes
Z
Z
1
1
1
2
cot
θ
dθ
=
csc2 θ − 1 dθ = − 3 (cot θ + θ) + C.
a3
a3
a
√
After substituting in cot θ = a/ x2 − a2 and θ = sec−1 (x/a), we get
Z
1
1
a
−1 x
dx = − 3 √
+ sec
+C
a
a
x(x2 − a2 )3/2
x2 − a2
2
on [a, ∞). Since the integrand is odd, we extend the result to an even
function and obtain
Z
1
1
a
−1 x dx = − 3 √
+ sec + C
a
a
x(x2 − a2 )3/2
x2 − a2
valid for x positive or negative.
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