7.2 notes calculus
Antidifferentiation by Substitution
Chapter 7 gives us some techniques for finding integrals or antiderivatives.
Definition: Indefinite Integral
The family of all antiderivatives of a function f(x) is the indefinite integral of f with respect to x and is
denoted by f ( x ) dx . If F is any function such that𝐹(𝑥) = 𝑓(𝑥), 𝑡ℎ𝑒 ∫ 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑥) + 𝐶, where C

is an arbitrary constant, called the constant of integration.

is the integral sign, the function f is the integrand of the integral and x is the variable of integration
When we find F(x) + C we have integrated f or evaluated the integral.
Example 1
Properties of Indefinite Integrals
∫ 𝑘 𝑓(𝑥)𝑑𝑥 = 𝑘 ∫ 𝑓(𝑥)𝑑𝑥 for any constant k
∫ (𝑓(𝑥) ± 𝑔(𝑥)) 𝑑𝑥 = ∫ 𝑓(𝑥)𝑑𝑥 ± ∫ 𝑔(𝑥)𝑑𝑥
Power Formulas
∫ 𝑢𝑛 𝑑𝑢 =
𝑢𝑛+1
𝑛+1
+ 𝐶, 𝑤ℎ𝑒𝑛 𝑛 ≠ −1
Trigonometric Formulas
∫ cos 𝑢 𝑑𝑢 = sin 𝑢 + 𝐶
∫ 𝑠𝑒𝑐 2 𝑢 𝑑𝑢 = tan 𝑢 + 𝐶
∫ sec(𝑢)tan(𝑢) 𝑑𝑢 = sec 𝑢 + 𝐶
1
∫ 𝑢−1 𝑑𝑢 = ∫ 𝑢 𝑑𝑢 = 𝑙𝑛|𝑢| + 𝐶
∫ sin 𝑢 𝑑𝑢 = −cos 𝑢 + 𝐶
∫ 𝑐𝑠𝑐 2 𝑢 𝑑𝑢 = −cot 𝑢 + 𝐶
∫ csc(𝑢)cot(𝑢) 𝑑𝑢 = −csc 𝑢 + 𝐶
Exponential and Logarithmic Formulas
∫ 𝑒𝑢 𝑑𝑢 = 𝑒𝑢 + 𝐶
∫ 𝑎𝑢 𝑑𝑢 =
∫ ln 𝑢 𝑑𝑢 = 𝑢 ln 𝑢 − 𝑢 + 𝐶
∫ log𝑎 𝑢 𝑑𝑢 = ∫
ln 𝑢
𝑢 ln(𝑢) − 𝑢
𝑑𝑢 =
+𝐶
ln 𝑎
ln 𝑎
Example 2 verifies Antiderivative formulas
Example 3 Paying Attention to the Differential
𝑎𝑢
ln 𝑎
+𝐶
Here is a table that might be helpful in antidifferentiating functions
Indefinite Integral
Reversed Derivative Formula
d  x n1 
n

  x , n  1
dx  n  1 
x n 1
 x dx  n  1  C , n  1
n
1
d
1
ln x 
dx
x
kx
d e
 ekx
dx k
d  cos kx 

  sin kx
dx 
k 
d  sin kx 

  cos kx
dx  k 
d
tan x  sec 2 x
dx
d
  cot x   csc2 x
dx
d
sec x  sec x tan x
dx
d
  csc x   csc x cot x
dx
 xdx  ln x  C
ekx
C
k
cos kx
 sin kx dx   k  C
sin kx
 cos kx dx  k  C
kx
 e dx 
 sec
 csc
2
x dx  tan x  C
2
x dx   cot x  C
 sec x tan x dx  sec x  C
 csc x cot x dx   csc x  C

Remember x n dx 
f
Example:
x n 1
C
n 1
F
x3
C
3
2
 x dx 
Recall also the power for derivatives could be used even if the base was a function by using the chain rule
5  2 x  1  2
d  2 x  1

dx
5
5
5
4
d u 
u n  du

dx n  1
dx
n 1
The inverse of the power rule and chain rule together is the power rule for integrals.
n
 u du 
u n 1
C
n 1
n  1
if n  1
then u n 
This formula is very helpful, but often hard to recognize.
1

u
1
 udu  ln u  C
x
Example:
3
 2  3x 2 dx To find this we need an antiderivative of the integrand. The integrand is a
4
product. This is difficult to integrate. However, if we integrate using a “u” substitution this becomes
easy.
x
3
 2  3 x 2 dx
4
x3  2 

u5
3
2
4
Let u  x  2 and du  3x dx   u du  
C
5
5
5
Check the answer by finding the derivative.
Do these problems:
  x  2

3
dx
4 x  1 dx
We do a “u” substitution when a function and its derivative (or a constant multiple of the derivative) make
up the integral.
 28  7 x  2  dx
3
#20
#17
 sin 3x dx
From #17 we can see that some trig functions aren’t too bad to integrate.
Example:
 sec
Example 4
Example 5
2
x dx
What about
 tan x dx
Since tan(x) isn’t in our integral table and it’s not the derivative of any function we know, we’ll have to
do something different. Remember that all functions can be written in terms of sines and cosines. Try
changing tangent to sin/cos
sin x
 tan x dx   cos xdx
What do you notice?
Answer: sin and cos are derivatives of one another
So we should do a “u” substitution. But, which one is u?
u  sin x
du  cos x dx
u  cos x
du   sin x dx
1
 u du
or
doesn ' t work
1
u
du   ln u  C   ln cos x  C
or ln
1
C
cos x
or
ln sec x  C
Substitution in Indefinite Integrals
 f ( g ( x))  g ( x) dx
= F(u) + C
=F(g(x)) + C
Example 7:

 f (u ) du
Substitute u=g(x), du=g′ (x) dx
Evaluate by finding an antiderivative F(u) of f (u)
Replace u by g(x)
Substitution in Definite Integrals
Substitute u = g (x), du = g′ (x) dx and integrate with respect to u from u = g (a) to u = g (b)

b
a
f ( g ( x))  g ( x) dx  
g (b )
g (a)
Example 8

f (u ) du

4
0
tan x sec2 x dx
What makes this problem different?
Continue on as normal with u = tan x and du = sec2 x dx


tan 
2


2 4
u 
1
1
4   tan 0 

4
0 u du   2   2  2  2  0  2
0
2

Or change the limits of integration by evaluating u at the limits.

1
u2 
1
1
4
tan 0 u du  0 u du   2   2  0  2
0
tan
1
It’s the same either way. You just have to decide if evaluating “u” at the limits saves effort. Usually it
does.
Example 9