FLC Math 400 Calculus I
5.5 Indefinite Integrals and the Substitution Rule
Page 1 of 4
By the FTC, differentiation and integration are inverse processes (adjusted by an integration constant). That is,
𝑑
𝑑
∫ 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑥) and ∫ 𝑑𝑥 𝑓(𝑥)𝑑𝑥 = 𝑓(𝑥) + 𝐶
𝑑𝑥
Note: In general, it is much more difficult to integrate than differentiate. In fact, all of chapter 8 is devoted to
techniques of integration. At this point, we don’t have a “Product Rule” for integrals and thus cannot integrate,
for example, ∫ 𝑥𝑒 𝑥 𝑑𝑥. (Can do this using Integration By Parts – Ch 8)
Theorem: Antidifferentiation of a Composite Function
Let 𝑔 be a function whose range is an interval 𝐼, and let 𝑓 be a function that is continuous on 𝐼. If
g is differentiable on its domain and 𝐹 is an antiderivative of 𝑓 on 𝐼, then
∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = 𝐹(𝑔(𝑥)) + 𝐶
If 𝑢 = 𝑔(𝑥), then 𝑑𝑢 = 𝑔′ (𝑥)𝑑𝑥 and
∫ 𝑓(𝑢)𝑑𝑢 = 𝐹(𝑢) + 𝐶
Def: Recall from sec. 5.1 that the set of all antiderivatives of the function 𝑓 is called the indefinite integral of 𝑓
with respect to 𝑥, and is symbolized by
∫ 𝑓(𝑥)𝑑𝑥
Rule If 𝑢 is any differentiable function, then ∫ 𝑢𝑛 𝑑𝑢 =
𝑢𝑛+1
𝑛+1
+ 𝐶, 𝑛 ≠ −1, 𝑛 𝑎𝑛𝑦 𝑛𝑢𝑚𝑏𝑒𝑟
Ex 1 (# 48) Find ∫ 4𝑥 3 sin(𝑥 4 ) 𝑑𝑥 by substituting 𝑢 = 𝑥 4 .
Often we will be asked to integrate functions that are the result of some chain rule. For cases like these
we employ a U substitution to help us find anti-derivatives of these types of functions.
Theorem The Substitution Rule
If 𝑢 = 𝑔(𝑥) is a differentiable function whose range is an interval 𝐼 and 𝑓 is continuous on 𝐼, then
∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = ∫ 𝑓(𝑢)𝑑𝑢
Ex 2 Integrate using substitution.
(# 11) ∫ 𝑥 3 (𝑥 4 + 3)2 𝑑𝑥 Hint: ∫ 𝑓(𝑔(𝑥))𝑔′ (𝑥)𝑑𝑥 = 𝐹(𝑔(𝑥)) + 𝐶 Which is 𝑓(𝑔(𝑥)) and which is 𝑔′ (𝑥)?
See bottom of first page of section 5.5 in book. “Exploration” Discuss whether or not we can find a pattern to
what we choose for our “U” and how we go about finding it each time.
FLC Math 400 Calculus I
5.5 Indefinite Integrals and the Substitution Rule
Page 2 of 4
Steps For 𝑈 −Subtitution:
1) Let 𝑢 be the most complicated part of the integrand. (Generally works.)
When deciding how to define 𝑢, look for what 𝑑𝑢 would be.
2) Find the differential 𝑑𝑢 and then solve for 𝑑𝑥.
3) Substitute in 𝑢 and 𝑑𝑢.
4) Integrate. Replace all 𝑢′𝑠 by 𝑥′𝑠.
Some common “tricks”
Sometimes 𝑔′ (𝑥) will be off by a multiple of a constant! If so, don’t worry, it will all work out.
a) (T# 8)∫ 12(𝑦 4 + 4𝑦 2 + 1)2 (𝑦 3 + 2𝑦)𝑑𝑦
Sometimes 𝑔(𝑥) & 𝑔′ (𝑥) will be hidden because a constant is added to 𝑔(𝑥). That is ok too, because when you
look for 𝑔′ (𝑥) you know that that constant will be gone.
3
b) (T# 20)∫
(1+√𝑥)
√𝑥
𝑑𝑥
c) ∫ 𝑡 3 (1 + 𝑡 4 )3 𝑑𝑡
Sometimes 𝑔(𝑥) & 𝑔′ (𝑥) are not obvious so you may need to try some different values for U before you find the
right one.
d) 5.5.5 ∫ tan2 𝑥 sec 2 𝑥 𝑑𝑥
FLC Math 400 Calculus I
5.5 Indefinite Integrals and the Substitution Rule
Page 3 of 4
This is one of my favorite tricks!! It looks so clever.
3
e) 5.5.94 ∫ 𝑡 ∙ √𝑡 − 4𝑑𝑡
f)
∫ 𝑥 3 √𝑥 2 + 1𝑑𝑥
g) (T# 40)∫(𝑠𝑖𝑛2𝜃)𝑒 sin
2𝜃
𝑑𝜃
h) (T# 42)∫
1 1/𝑥
𝑒 sec(1 +
𝑥2
i)
ln √𝑡
𝑑𝑡
𝑡
(T# 44)∫
𝑒 1/𝑥 ) tan( 1 + 𝑒 1/𝑥 )𝑑𝑥
Ex 3 Find ∫ sin2 𝑥 𝑑𝑥 using the trig identity: sin2 𝑥 =
1−cos 2𝑥
.
2
𝑥 sin 2𝑥
−
+𝐶
2
4
𝑥 sin 2𝑥
∫ cos 2 𝑥 𝑑𝑥 = +
+𝐶
2
4
∫ sin2 𝑥 𝑑𝑥 =
FLC Math 400 Calculus I
5.5 Indefinite Integrals and the Substitution Rule
Page 4 of 4
Definite Integrals and U subs
When you make a u substitution on a definite integral, it is like translating the problem from one language to
another. With a definite integral, all you have to remember is that your limits of integration are in the original
language and if you make a u substitution, you need to translate the limits as well!!
Ex: Evaluate the definite Integral:
4
a) 5.5.96 ∫−2 𝑥 2 (𝑥 3 + 8)2 𝑑𝑥
√2
b) 5.5.104 ∫0 𝑥𝑒 −(𝑥
2 /2)
𝑑𝑥
Integration of Even and Odd Functions
Integration of Even and Odd Functions
Let 𝑓 be integrable on the closed interval [−𝑎, 𝑎]
1. If 𝑓 is an even function, then
𝑎
𝑎
∫ 𝑓(𝑥)𝑑𝑥 = 2 ∫ 𝑓(𝑥)𝑑𝑥
−𝑎
0
2. If 𝑓 is an odd function, then
𝑎
∫ 𝑓(𝑥)𝑑𝑥 = 0
−𝑎
Ex: Evaluate the integral:
𝜋/2
a) ∫–𝜋/2 𝑆𝑖𝑛2 𝑥𝐶𝑜𝑠𝑥𝑑𝑥
𝜋/4
b) ∫–𝜋/4 𝑆𝑖𝑛𝑥𝐶𝑜𝑠𝑥𝑑𝑥