Methods of Integration
Professor D. Olles
January 28, 2014
1
Substitution
The derivative of a composition of functions can be found using the chain rule
form
d
[f (g(x))] = f 0 (g(x)) g 0 (x)
dx
Rewriting the derivative form in the differential form give us
d [f (g(x))] = f 0 (g(x)) g 0 (x)dx
Integrating both sides of the equation results in
Z
Z
d [f (g(x))] = f 0 (g(x)) g 0 (x)dx
Z
f (g(x)) + C =
Let
Then
f 0 (g(x)) g 0 (x)dx
u = g(x)
= g 0 (x)
du = g 0 (x)dx
du
dx
The above equation can be written as
Z
f 0 (u)du = f (u) + C
Or, if the integrand is some function y = f (g(x)),
Z
f (u)du = F (u) + C
Note that in the case of the definite integral, it is necessary to convert our limits
of integration to match up with the new independent variable. That is:
if x = a then u = g(a)
if x = b then u = g(b)
1
and the integral formula becomes:
Z b
Z g(b)
g(b)
f (g(x))g 0 (x)dx =
f (u)du = F (u)|g(a) = F (g(b)) − F (g(a))
a
g(a)
Example 1.1 Evaluate the following indefinite integral using the method of
substitution.
Z
e2x dx
Solution.
u = 2x
du = 2dx
1
du = dx
2
Z
Z
1
2x
e dx = eu du
2
Z
1
eu du
=
2
1
= eu + C
2
1 2x
= e +C
2
J
Example 1.2 Evaluate the following indefinite integral using the method of
substitution.
Z
x2 + 1
dx
x3 + 3x
Solution.
u = x3 + 3x
du = (3x2 + 3)dx = 3(x2 + 1)dx
1
du = (x2 + 1)dx
3
Z
Z
x2 + 1
11
dx
=
du
3
x + 3x
u3
Z
1
1
=
du
3
u
1
= ln |u| + C
3
1
= ln |x3 + 3x| + C
3
J
2
Example 1.3 Evaluate the following indefinite integral using the method of
substitution.
Z arctan x
e
dx
1 + x2
Solution.
u = arctan x
1
du =
dx
1 + x2
Z arctan x
Z
e
dx = eu du
1 + x2
= eu + C
= earctan x + C
J
Example 1.4 Evaluate the following definite integral using the method of
substitution.
Z e
sin (ln x)
dx
x
1
Solution.
u = ln x
1
du = dx
x
x = 1 =⇒ u = ln 1 = 0
x = e =⇒ u = ln e = 1
Z 1
sin (ln x)
dx =
sin udu
x
1
0
Z
e
1
= − cos u|0
= − cos (1) − − cos (0)
= − cos (1) + 1
= 1 − cos (1)
J
3
2
Integration by Parts
The derivative of the product of two function can be found using the form
d
[f (x)g(x)] = f 0 (x)g(x) + f (x)g 0 (x)
dx
If we rewrite the derivative form in the differential form, we have
d [f (x)g(x)] = [f 0 (x)g(x) + f (x)g 0 (x)] dx
Integrating both sides allows us to solve this differential equation.
Z
Z
d [f (x)g(x)] = [f 0 (x)g(x) + f (x)g 0 (x)]dx
Z
f (x)g(x) =
f 0 (x)g(x)dx +
Z
f (x)g 0 (x)dx
Solving for the second term on the right hand side yields
Z
Z
0
f (x)g (x)dx = f (x)g(x) − f 0 (x)g(x)dx
Let
Then
u = f (x)
du = f 0 (x)dx
dv = g 0 (x)dx
v = g(x)
and
and
Our equation above can be written in simpler form. That is
Z
Z
udv = uv − vdu
Z
f (x)g 0 (x)dx = f (x)g(x) −
Z
f 0 (x)g(x)dx
In the case of the definite integral, we have
Z
a
b
b
f (x)g 0 (x)dx = f (x)g(x)|a −
Z
b
f 0 (x)g(x)dx
a
Note that since we are not actually performing a change of variables, it is not
necessary to change our limits of integration.
4
Example 2.1
Evaluate the following integral using Integration by Parts.
Z
x2 e2x dx
Solution.
u1 = x2
dv1 = e2x dx
du1 = 2xdx
v1 = 21 e2x
Z Z
1 2x
1 2x
2 2x
2
x e dx = x
e
e
−
2xdx
2
2
Z
1
= x2 e2x − xe2x dx
2
u2 = x
dv2 = e2x dx
du2 = dx
v2 = 21 e2x
Z
1
1
1 2x
= x2 e2x − xe2x −
e dx
2
2
2
=
1 2 2x 1 2x 1 2x
x e − xe + e + C
2
2
4
J
Example 2.2
Evaluate the following integral using Integration by Parts.
Z
e3z cos 3zdz
Solution.
w = 3z
dw = 3dz
1
dw = dz
3
Z
Z
1
e3z cos 3zdz =
ew cos wdw
3
u1 = ew
du1 = ew dw
Z
e
3z
1
cos 3zdz =
3
Z
dv1 = cos wdw
v1 = sin w
Z
1 w
w
e cos wdw =
e sin w − e sin wdw
3
w
u2 = ew
du2 = ew dw
dv2 = sin wdw
v2 = − cos w
5
1
3
Z
Z
1 w
e sin w − −ew cos w − ew (− cos w)dw
3
Z
Z
1
1
1
1
ew cos wdw = ew sin w + ew cos w −
ew cos wdw
3
3
3
3
Z
2
1
1
ew cos wdw = ew sin w + ew cos w + C
3
3
3
Z
1
1
1
ew cos wdw = ew sin w + ew cos w + C
3
6
6
Z
1
1
e3 z cos 3zdz = e3 z sin 3z + e3 z cos 3z + C
6
6
ew cos wdw =
J
Evaluate the following integral using Integration by Parts.
Z
x2 arctan xdx
Example 2.3
Solution.
dv = x2 dx
3
v = x3
Z
Z
1
1
x3
x2 arctan xdx = x3 arctan x −
dx
3
3
1 + x2
u = arctan x
1
du = 1+x
2 dx
u = 1 + x2
du = 2xdx
1
du = xdx
2
x2
du = x3 dx
2
u−1
du = x3 dx
2
Z
Z
1 3
1
1u−1
x arctan xdx = x arctan x −
du
3
3
u 2
Z 1
1
1
1−
du
= x3 arctan x −
3
6
u
2
1
1 3
x arctan x − [u − ln |u|] + C
3
6
1
1
= x3 arctan x −
1 + x2 − ln (1 + x2 ) + C
3
6
=
6
=
1 3
1 1
1
x arctan x − + x2 − ln (1 + x2 ) + C
3
6 6
6
1 2 1
1 3
= x arctan x + x − ln (1 + x2 ) + C
3
6
6
J
Example 2.4
Evaluate the following integral using Integration by Parts.
Z
ln x
dx
x2
Solution.
dv = x−2 dx
v = − x1
Z
Z
ln x
ln x
1
dx
=
−
+
dx
2
x
x
x2
u = ln x
du = x1 dx
=−
ln x 1
− +C
x
x
J
Example 2.5
Evaluate the following integral using Integration by Parts.
Z e
x2 ln xdx
1
Solution.
u = ln x
du = x1 dx
dv = x2 dx
3
v = x3
e
Z
1 e 2
x3 ln x −
x dx
=
3 1 3 1
e
x3 e3
=
−0−
3
9 1
=
e3
e3
1
−
+
3
9
9
=
2e3 + 1
9
J
7
3
3.1
Trigonometric Substitution
Utilizing the Pythagorean Identity 1 − sin2 θ = cos2 θ
Consider the Pythagorean Identity sin2 θ + cos2 θ = 1 and its alternate form
1 − sin2 θ = cos2 θ
If we choose to multiply both sides of the equation by a perfect square constant
a2 we have
a2 (1 − sin2 θ) = a2 cos2 θ
a2 − a2 sin2 θ = a2 cos2 θ
Since the RHS is a product of perfect squares, taking the square root of both
sides would reduce the power simply.
p
√
a2 − a2 sin2 θ = a2 cos2 θ
p
a2 − a2 sin2 θ = a cos θ
If we let x = a sin θ, then x2 = a2 sin2 θ and
p
a2 − x2 = a cos θ
So, the substitution we should make to change our variables is
x = a sin θ
dx = a cos θdθ
The resulting integrand will be a function of θ yielding an antiderivative as a
function of θ. It is then necessary to perform a backsubstitution in order to write
our answer in terms of the independent variable x. We do this by returning to
the claim that
x = a sin θ
Then
x
OP P
=
a
HY P
and all other 5 trigonometric evaluations can be obtained from the resulting
triangle below.
sin θ =
8
Example 3.1
Evaluate the following integral.
Z
x
√
dx
4 − x2
Solution.
x = 2 sin θ
dx = 2 cos θdθ
Z
=
2 sin θ
p
2 cos θdθ
4 − (2 sin θ)2
Z
=
Z
=
2 sin θ
p
4 − 4 sin2 θ
2 sin θ
2 cos θdθ
p
2 cos θdθ
4(1 − sin2 θ)
Z
2 sin θ
√
=
2 cos θdθ
4 cos2 θ
Z
2 sin θ
=
2 cos θdθ
2 cos θ
Z
= 2 sin θdθ
= −2 cos θ + C
x = 2 sin θ =⇒ sin θ =
√
cos θ =
x
2
4 − x2
2
√
4 − x2
= −2
+C
2
p
= − 4 − x2 + C
J
9
3.2
Utilizing the Pythagorean Identity sec2 θ − 1 = tan2 θ
Consider the Pythagorean Identity sin2 θ + cos2 θ = 1 and its alternate form
sec2 θ − 1 = tan2 θ
If we choose to multiply both sides of the equation by a perfect square constant
a2 we have
a2 (sec2 θ − 1) = a2 tan2 θ
a2 sec2 θ − a2 = a2 tan2 θ
Since the RHS is a product of perfect squares, taking the square root of both
sides would reduce the power simply.
p
p
a2 sec2 θ − a2 = a2 tan2 θ
p
a2 sec2 θ − a2 = a tan θ
If we let x = a sec θ, then x2 = a2 sec2 θ and
p
x2 − a2 = a tan θ
So, the substitution we should make to change our variables is
x = a sec θ
dx = a sec θ tan θdθ
The resulting integrand will be a function of θ yielding an antiderivative as a
function of θ. It is then necessary to perform a backsubstitution in order to write
our answer in terms of the independent variable x. We do this by returning to
the claim that
x = a sec θ
Then
x
HY P
=
a
ADJ
and all other 5 trigonometric evaluations can be obtained from the resulting
triangle below.
sec θ =
10
Example 3.2
Evaluate the following integral.
Z √ 2
4x − 9
dx
x2
Solution.
Z p
=
(2x)2 − 9
dx
x2
1
u=x
2
du = 2dx
1
du = dx
2
Z √ 2
u −91
=
2 du
1
2
2u
√
Z
1
u2 − 9
du
=
1 2
2
4u
Z √ 2
u −9
du
=2
u2
u = 2x =⇒
u = 3 sec θ
du = 3 sec θ tan θdθ
Z p
(3 sec θ)2 − 9
=2
3 sec θ tan θdθ
(3 sec θ)2
Z √
9 sec2 θ − 9
=2
3 sec θ tan θdθ
9 sec2 θ
Z p
9(sec2 θ − 1)
=2
tan θdθ
3 sec θ
Z √
9 tan2 θ
=2
tan θdθ
3 sec θ
Z
3 tan θ
=2
tan θdθ
3 sec θ
Z
tan2 θ
=2
dθ
sec θ
Z
sin2 θ
=2
cos θdθ
cos2 θ
11
sin2 θ
dθ
cos θ
Z
=2
1 − cos2 θ
dθ
cos θ
Z 1
− cos θ dθ
=2
cos θ
Z
= 2 (sec θ − cos θ)dθ
Z
=2
= 2 [ln | sec θ + tan θ| − sin θ] + C
u = 3 sec θ =⇒ sec θ =
√
tan θ =
√
u
3
u2 − 9
3
u2 − 9
u
"
#
u √u2 − 9 √u2 − 9
= 2 ln +
+C
−
3
3
u
" #
2x √4x2 − 9 √4x2 − 9
+
+C
= 2 ln −
3
3
2x
2x √4x2 − 9 √4x2 − 9
= 2 ln +
+C
−
3
3
x
sin θ =
J
12
3.3
Utilizing the Pythagorean Identity 1 + tan2 θ = sec2 θ
Consider the Pythagorean Identity sin2 θ + cos2 θ = 1 and its alternate form
1 + tan2 θ = sec2 θ
If we choose to multiply both sides of the equation by a perfect square constant
a2 we have
a2 (1 + tan2 θ) = a2 sec2 θ
a2 + a2 tan2 θ = a2 sec2 θ
Since the RHS is a product of perfect squares, taking the square root of both
sides would reduce the power simply.
p
√
a2 + a2 tan2 θ = a2 sec2 θ
p
a2 + a2 tan2 θ = a sec θ
If we let x = a tan θ, then x2 = a2 tan2 θ and
p
a2 + x2 = a sec θ
So, the substitution we should make to change our variables is
x = a tan θ
dx = a sec2 θdθ
The resulting integrand will be a function of θ yielding an antiderivative as a
function of θ. It is then necessary to perform a backsubstitution in order to write
our answer in terms of the independent variable x. We do this by returning to
the claim that
x = a tan θ
Then
x
OP P
=
a
ADJ
and all other 5 trigonometric evaluations can be obtained from the resulting
triangle below.
tan θ =
13
Example 3.3
Evaluate the following integral.
Z
x2
√
dx
9 + x2
Solution.
x = 3 tan θ
dx = 3 sec2 θdθ
(3 tan θ)2
Z
=
p
9 + (3 tan θ)2
9 tan2 θ
p
3 sec2 θdθ
9 + 9 tan2 θ
Z
=
9 tan2 θ
Z
=
3 sec2 θdθ
p
2
9(1 + tan θ)
3 sec2 θdθ
9 tan2 θ
√
3 sec2 θdθ
9 sec2 θ
Z
=
9 tan2 θ
3 sec2 θdθ
3 sec θ
Z
=
Z
=
9 tan2 θ sec θdθ
Z
=9
tan θ sec θ tan θdθ
u = tan θ
dv = sec θ tan θdθ
du = sec2 θdθ
v = sec θ
Z
= 9 sec θ tan θ − sec3 θdθ
1
1
= 9 sec θ tan θ −
sec θ tan θ + ln | sec θ + tan θ| + C
2
2
=
9
9
sec θ tan θ − ln | sec θ + tan θ| + C
2
2
x = 3 tan θ =⇒ tan θ =
14
x
3
√
9 + x2
3
√
!
√
9 + x2 x 9 9 + x2
9
x =
+ +C
− ln 2
3
3
2 3
3
√
9 9 + x2 + x xp
9 + x2 − ln =
+C
2
2 3
sec θ =
J
15
4
Partial Fraction Decomposition & Polynomial
Long Division
(x)
When integration rational functions of the form R(x) = N
D(x) , where N and
D are polynomials of degree m and n respectively, it may be necessary to perform some algebraic manipulation to rewrite the function in a simpler form, or
one whose antiderivative is easier to recognize. That is, of course, if any other
method has been exhausted.
Consider rational functions whose antiderivatives are not recognizable immediately and cannot be integrated by means of substitution. Let
R(x) =
4.1
am xm + am−1 xm−1 + · · · + a2 x2 + a1 x + a0
N (x)
=
D(x)
bn xn + bn−1 xn−1 + · · · + b2 x2 + b1 x + b0
m ≥ n Long Division of Polynomials
Let us review the alternate form division algorithm
r(x)
n(x)
= q(x) +
d(x)
d(x)
where n(x) is the numerator of the rational expression, d(x) is the denominator,
q(x) is the resulting quotient and r(x) is any remainder as a function of x whose
degree is less than the degree of the denominator. We will use an example to
review this concept.
Example 4.1
Evaluate the following integral.
Z
x2
dx
x+1
Solution.
x−1
x+1
2
x
− x2 − x
−x
x+1
1
Z
1
x2
dx =
x−1+
dx
x+1
x+1
Z
Z
1
= (x − 1)dx +
dx
x+1
Z u=x+1
du = dx
16
Z
x2
1
−x+
du
2
u
x2
=
− x + ln |u| + C
2
x2
− x + ln |x + 1| + C
=
2
=
J
4.2
4.2.1
m < n Partial Fraction Decomposition
Unique Linear Factors
If the denominator is the product of unique linear factors, we may rewrite the
rational expression as
A
B
C
N (x)
=
+
+ ··· +
(ax + b)(cx + d) · · · (ex + f )
ax + b cx + d
ex + f
where A, B, ..., C are undetermined constants. We may then algebraically solve
for these constants.
Example 4.2
Z
x2
x+1
dx
+ 5x + 6
Solution.
x+1
A
B
x+1
=
=
+
x2 + 5x + 6
(x + 3)(x + 2)
x+3 x+2
x+1
A(x + 2) + B(x + 3)
=
(x + 3)(x + 2)
(x + 3)(x + 2)
x + 1 = Ax + 2A + Bx + 3B
x + 1 = (A + B)x + (2A + 3B)
1=A+B
1−A=B
1 = 2A + 3B
1 = 2A + 3(1 − A)
1 = 2A + 3 − 3A
−2 = −A
A=2
1−2=B
B = −1
2
1
x+1
=
−
x2 + 5x + 6
x+3 x+2
Z
Z x+1
2
1
dx
=
−
dx
x2 + 5x + 6
x+3 x+2
= 2 ln |x + 3| − ln |x + 2| + C
J
17
4.2.2
Repeated Linear Factors
If the denominator is the product of repeated linear factors, we may rewrite the
rational expression as
A
C
N (x)
B
=
+ ··· +
+
(ax + b)n
ax + b (ax + b)2
(ax + b)n
where A, B, ..., C are undetermined constants. We may then algebraically solve
for these constants.
Example 4.3
Z
Solution.
x2
x−3
dx
x2 + 4x + 4
x−3
B
x−3
A
=
+
=
2
+ 4x + 4
(x + 2)
x + 2 (x + 2)2
x−3
A(x + 2) + B
=
2
(x + 2)
(x + 2)2
x − 3 = Ax + 2A + B
1=A
−3 = 2A + B
−3 = 2(1) + B
−5 = B
x−3
1
5
=
−
x2 + 4x + 4
x + 2 (x + 2)2
Z
Z x−3
1
5
dx
=
−
dx
x2 + 4x + 4
x + 2 (x + 2)2
= ln |x + 2| +
5
+C
x+2
J
4.2.3
Unique Quadratic Factors
If the denominator is the product of unique quadratic factors, we may rewrite
the rational expression as
Ax + B
Cx + D
Ex + F
N (x)
= 2
+
+· · ·+ 2
(ax2 + bx + c)(dx2 + ex + f ) · · · (gx2 + hx + i)
ax + bx + c dx2 + ex + f
gx + hx + i
Example 4.4
Z
x2 − 3x + 5
dx
(x2 + 1)(x2 + 4)
18
Solution.
x2 − 3x + 5
Ax + B
Cx + D
= 2
+ 2
2
+ 1)(x + 4)
x +1
x +4
(x2
x2 − 3x + 5
(Ax + B)(x2 + 4) + (Cx + D)(x2 + 1)
=
(x2 + 1)(x2 + 4)
(x2 + 1)(x2 + 4)
x2 − 3x + 5 = Ax3 + Bx2 + 4Ax + 4B + Cx3 + Dx2 + Cx + D
x2 − 3x + 5 = (A + C)x3 + (B + D)x2 + (4A + C)x + (4B + D)
A+C =0
A = −C
A = −1
B+D =1
B =1−D
B=
4A + C = −3
4B + D = 5
−4C + C = −3
−3C = −3
C=1
4(1 − D) + D = 5
4 − 3D = 5
D = − 31
4
3
−x + 34
x − 13
x2 − 3x + 5
=
+
(x2 + 1)(x2 + 4)
x2 + 1
x2 + 4
Z
Z
x2 − 3x + 5
x
4/3
x
1/3
dx
=
−
+
+
−
dx
(x2 + 1)(x2 + 4)
x2 + 1 x2 + 1 x2 + 4 x2 + 4
Z
Z
Z
Z
4
1
1
x
1
x
dx
+
dx
+
dx
−
dx
=−
2
2
2
2
x +1
3
x +1
x +4
3
x +4
u = x2 + 1
w = x2 + 4
x = 2 tan θ
u = 2xdx
dw = 2xdx dx = 2 sec2 θdθ
1
1
2 du = dx
2 dw = dx
Z
Z
Z
Z
1
4
1
1
1
1
1
1
du +
dx
+
dw
−
2 sec2 θdθ
=−
2
2
u
3
x +1
2
w
3
(2 tan θ)2 + 4
Z
1
4
1
1
= − ln |u| + arctan x + ln |w| −
dθ
2
3
2
6
1
4
1
1
= − ln (x2 + 1) + arctan x + ln (x2 + 4) − θ + C
2
3
2
6
x
1
4
1
1
= − ln (x2 + 1) + arctan x + ln (x2 + 4) − arctan
+C
2
3
2
6
2
J
19
4.2.4
Repeated Quadratic Factors
If the denominator is the product of repeated quadratic factors, we may rewrite
the rational expression as
Ax + B
Ex + F
Cx + D
N (x)
= 2
+ ··· +
+
(ax2 + bx + c)n
ax + bx + c (ax2 + bx + c)2
(ax2 + bx + c)n
Example 4.5
Z
Solution.
x2 − 3x + 5
dx
(x2 + 1)2
x2 − 3x + 5
Ax + B
Cx + D
= 2
+ 2
(x2 + 1)2
x +1
(x + 1)2
x2 − 3x + 5
(Ax + B)(x2 + 1) + Cx + D
=
2
2
(x + 1)
(x2 + 1)2
x2 − 3x + 5 = Ax3 + Bx2 + Ax + B + Cx + D
x2 − 3x + 5 = Ax3 + Bx2 + (A + C)x + (B + D)
A=0
B=1
A + C = −3
C = −3
B+D =5
1+D =5
D=4
x2 − 3x + 5
1
−3x + 4
= 2
+
(x2 + 1)2
x + 1 (x2 + 1)2
Z 2
Z x − 3x + 5
1
−3x + 4
dx
=
+
dx
(x2 + 1)2
x2 + 1 (x2 + 1)2
Z
Z
Z
x
1
1
dx − 3
dx + 4
dx
=
2
2
2
2
x +1
(x + 1)
(x + 1)2
u = x2 + 1
x = tan θ
du = 2xdx dx = sec2 θdθ
1
2 du = dx
Z
Z
1
1
1
sec2 θdθ
= arctan x − 3
du
+
4
2
2
2
u
(tan θ + 1)2
Z
3
= arctan x +
+ 4 cos2 θdθ
2u
Z
3
= arctan x +
+ 2 (1 + cos 2θ)dθ
2(x2 + 1)
3
1
= arctan x +
+
2
θ
+
sin
2θ
+C
2(x2 + 1)
2
= arctan x +
3
+ 2 [θ + sin θ cos θ] + C
2(x2 + 1)
20
= arctan x +
3
x+1
+C
+ 2 arctan x + 2 √
+ 1)
x2 + 1
2(x2
= 3 arctan x +
3
2x + 2
+√
+C
2(x2 + 1)
x2 + 1
J
21