LECTURE 15: SUBSTITUTION RULE
MINGFENG ZHAO
February 06, 2015
Theorem 1.
Z
p
x dx
=
Z
cos(ax) dx
=
sin(ax) dx
=
sec2 (ax) dx
=
csc2 (ax) dx
Z
eax dx
=
Z
Z
Z
Z
1
dx
a2 − x2
Z
1
dx
x2 + a2
√
=
=
=


xp+1
p+1
+ C,
if p 6= −1
 ln |x| + C,
if p = −1.
1
sin(ax) + C
a
1
− cos(ax) + C
a
1
tan(ax) + C
a
1
− cot(ax) + C
a
1 ax
e +C
a
x
sin−1
+C
a
x
1
tan−1
+C
a
a
Theorem 2 (Fundamental theorem of calculus). Let f be continuous, and F be an antiderivative of f , then
Z b
b
f (x) dx = F (b) − F (a) := F (x)|a .
a
x
Z
Moreover, let A(x) :=
f (t) dt, then A is an antiderivative of f , that is,
a
A0 (x) = f (x).
Z
1
4
Example 1. Compute
1
16
√
t−1
dt.
t
Since
Z √
t−1
dt
t
Z =
1
1
1
√ −
t
t
dt
2
MINGFENG ZHAO
Z
=
− 21
t
Z
dt −
1
dt
t
=
1 − 1 +1
t 2 − ln |t| + C
1 − 21
=
2t 2 − ln |t| + C.
1
Then we get
Z
1
4
√
1
16
t−1
dt
t
=
14
1
2t 2 − ln |t| 1
16
=
12
12
1
1
1
1
2·
− ln − 2 ·
+ ln
4
4
16
16
1
1
+ ln 4 − 2 · − ln 16
2
4
1
= 1 + ln 4 − − ln 16
2
1
+ ln 4 − 2 ln 4
=
2
1
=
− ln 4.
2
=
2·
Theorem 3. Let f be continuous, a(x) and b(x) be differentiable, then
d
dx
d
Example 2. Compute
dx
In fact, we have
Z
Z
b(x)
f (t) dt = f (b(x))b0 (x) − f (a(x))a0 (x).
a(x)
x2
cos(t2 ) dt.
0
d
dx
Z
x2
cos(t2 ) dt
d 2
d
(x ) − cos(02 ) ·
(0)
dx
dx
=
cos(x2 ) ·
=
cos(x2 ) · 2x
=
2x cos(x2 ).
0
Substitution rule for indefinite integrals
Recall the chain rule, we have
d
f (g(x)) = f 0 (g(x))g 0 (x).
dx
LECTURE 15: SUBSTITUTION RULE
3
Theorem 4. Let u = g(x) be differentiable, then
Z
0
f (g(x))g (x) dx =
Z
f (u) du.
Procedure of Substitution Rule or Change of Variables or u-Substitution:
I. Given an indefinite integral involving a composition function f (g(x)), identify an inner function u = g(x) such
that a constant multiple of g 0 (x) appears in the integral.
1
du) in the integral.
II. Substitute u = g(x) and du = g 0 (x)dx (that is, dx = 0
g (x)
III. Evaluate the new indefinite integral with respect to u.
IV. Write the result in terms of x using u = g(x).
Z
Example 3. Compute
(2x + 1)3 dx.
Let u = 2x + 1, then du = 2dx, that is, dx =
Z
Z
Example 4. Compute
(2x + 1)3 dx
e10x dx.
1
du. Then
10
Z
1
dx =
eu ·
du Let u = 10x
10
Z
1
=
eu du
10
1 u
=
e +C
10
1 10x
e
+ C Since u = 10x.
=
10
Let u = 10x, then du = 10dx, that is, dx =
Z
Z
Example 5. Compute
1
du. Then
2
Z
1
=
u3 · du Let u = 2x + 1
2
Z
1
=
u3 du
2
1
1
=
·
u3+1 + C
2 1+3
1 4
u +C
=
8
1
=
(2x + 1)4 + C Since u = 2x + 1.
8
e10x
x4 (x5 + 6)9 dx.
4
MINGFENG ZHAO
Let u = x5 + 6, then du = 5x4 dx, that is, x4 dx = 15 du. Then
Z
Z
1
4 5
9
x (x + 6) dx =
u9 · du Let u = x5 + 6
5
Z
1
u9 du
=
5
1
1
=
·
u1+9 + C
5 1+9
1 10
=
u +C
50
1 5
=
(x + 6)10 + C Since u = x5 + 6.
50
Z
Example 6. Compute
cos3 (x) sin(x) dx.
Let u = cos(x), then du = − sin(x)dx, that is, sin(x)dx = −du. Then
Z
Z
3
cos (x) sin(x) dx = − u3 du Let u = cos(x)
=
=
=
Z
1
u3+1 + C
1+3
1
− u4 + C
4
1
− cos4 (x) + C Since u = cos(x).
4
−
x
dx.
x+1
Approach I: Let u = x + 1, then x = u − 1 and du = dx. Then
Z
Z
x
u−1
√
√ du Let u = x + 1
dx =
u
x+1
Z √
1
u− √
=
du
u
Z
Z
1
1
=
u 2 du − u− 2 du
√
Example 7. Compute
=
=
=
Approach II: Let u =
√
1
1
1
1− 12
2 +1 −
+C
1u
1u
1+ 2
1− 2
1
2 3
u 2 − 2u 2 + C
3
3
1
2
(x + 1) 2 − 2(x + 1) 2 + C
3
Since u = x + 1.
x + 1, then x = u2 − 1 and dx = 2udu. Then
Z
Z 2
√
x
u −1
√
dx =
· 2u du Let u = x + 1
u
x+1
LECTURE 15: SUBSTITUTION RULE
5
Z
=
=
=
=
(u2 − 1) du
2 3
2
u −u +C
3
2
4 3
u − 2u + C
3
3
1
2
(x + 1) 2 − 2(x + 1) 2 + C
3
Since u =
√
x + 1.
Substitution rule for definite integrals
Theorem 5. Let u = g(x) be differentiable, then
Z b
Z
f (g(x))g 0 (x) dx =
a
g(b)
f (u) du.
g(a)
Department of Mathematics, The University of British Columbia, Room 121, 1984 Mathematics Road, Vancouver, B.C.
Canada V6T 1Z2
E-mail address: mingfeng@math.ubc.ca