Calculus- Early Transcendentals, 2021a

7.3. Trigonometric Substitutions 271
∫ √ 25x 2 − 4
x
dx =
=
√
∫ 25 4sec2 θ
25
− 4
2
5 secθ 2
5
secθ tanθ dθ Using the substitution
∫ √
4(sec 2 θ − 1) · tanθ dθ Cancelling
2∫ √
= tan 2 θ · tanθ dθ
∫
Using tan 2 θ + 1 = sec 2 θ
= 2
∫
tan 2 θ dθ
Simplifying
= 2 (sec 2 θ − 1)dθ
Using tan 2 θ + 1 = sec 2 θ
= 2(tanθ − θ)+C Since ∫ sec 2 θ dθ = tanθ +C
For tanθ, we use a right triangle.
x = 2 5 secθ → x = 2 1
5 cosθ
Using SOH CAH TOA, the triangle is then
→
cosθ = 2 5x
For θ by itself, we use θ = sec −1 (5x/2). Thus,
∫ √ (√
25x 2 − 4 25x
dx = 2
2 − 4
x
2
( ) ) 5x
− sec −1 +C
2
In the context of the previous example, some resources give alternate guidelines when choosing a
trigonometric substitution.
√
a 2 − b 2 x 2 → x = a b sinθ
♣
√
b 2 x 2 + a 2 or (b 2 x 2 + a 2 ) → x = a b tanθ
√
b 2 x 2 − a 2 → x = a b secθ
We next look at a tangent substitution.