MATH 1020 WORKSHEET 5.3-5.5
Basic Integration and u-Substitution
Evaluate the following integrals.
Z
u2 + eu du
Solution. Using basic integration rules one writes
Z
u2 + eu du =
u3
+ eu + C
3
Thus
Z
Z
u2 + eu du =
u3
+ eu + C.
3
2x cos (x2 ) dx
Solution. Using u-substitution with u = x2 and du = 2x dx one writes
Z
Z
2
2x cos (x ) dx
=
Z
cos (x2 ) 2x dx
cos (u) du
= sin (u) + C
= sin(x2 ) + C
Thus
Z
2x cos (x2 ) dx = sin(x2 ) + C .
Z
ln 2
ex dx
0
Solution. Using basic integration rules one writes
Z
0
ln 2
2
ex dx = ex |ln
0
= eln 2 − e0
= 2−1
= 1
Thus the definite integral gives 1 .
MATH 1020 WORKSHEET 7.1
Integration by Parts
Z
u dv = uv −
Z
v du
select u using LIATE.
L=Logarithmic, I=Inverses, A=Algebraic(or use P=Polynomials), T=Trig, E=Exponential.
Z
x
dx
ex
Solution. Using integration by parts formula one writes
Z
Z
x
dx =
x e−x dx
ex
= −xe−x −
= −xe
−x
+
Z
(−e−x ) dx
Z
e−x dx
= −xe−x − e−x + C
Thus
Z
Z
x
dx = −xe−x − e−x + C .
ex
ex cos (2x) dx
Solution. Using twice the integration by parts formula one writes
Z
ex cos (2x) dx =
Z
cos (2x) ex dx
= ex cos (2x) − 2
x
= e cos (2x) + 2
Z
ex (−2 sin (2x)) dx
Z
sin (2x) ex dx
= ex cos (2x) + 2[ex sin (2x) −
x
x
Z
= e cos (2x) + 2e sin (2x) − 4
Noting that the term
5
Z
R
ex (2 cos (2x))] dx
Z
ex cos (2x) dx.
ex cos x dx appears on both sides of the equation, one writes
ex cos (2x) dx = ex cos (2x) + 2ex sin (2x) + C.
Thus
Z
1
ex cos (2x)) dx = [ex cos (2x) + 2ex sin (2x)] + C .
5
Z
1
ln (x2 + 1) dx
0
Solution. Using integration by parts formula one writes
Z
1
2
ln (x + 1) dx =
Z
0
1
ln (x2 + 1) · (1) dx
0
=
Z
1
ln (x2 + 1) 1 dx
0
1
2x
x 2
dx
0
x +!
1
0
Z 1
1
2x2
dx
x ln (x2 + 1) −
0
x2 + 1
0
!
Z 1
2
1
(2x
+
2)
−
2
x ln (x2 + 1) −
dx
0
x2 + 1
0
!
Z 1
1
2 (x2 + 1) − 2
2
x ln (x + 1) −
dx
0
x2 + 1
0
Z 1
1
2
2− 2
x ln (x2 + 1) −
dx
0
x +1
0
Z
1
x ln (x + 1) −
2
=
=
=
=
=
1
= x ln (x2 + 1) − 2x|10 + 2 arctan (x)|10
0
= ( 1 ln (2) − 0 ) − ( 2 − 0 ) + ( 2 arctan (1) − 2 arctan (0) )
π
= ln (2) − 2 + 2 − 2 · 0
4
π
= ln (2) − 2 + .
2
Thus
Z
0
1
ln (x2 + 1) dx = ln (2) − 2 +
π
.
2
MATH 1020 WORKSHEET 7.2
Trigonometric Integrals
If possible replace cos2 x with 1 − sin2 x OR sin2 x with 1 − cos2 x in order to use
1
u-substitution. Otherwise, use the identities cos2 x = (1 + cos (2x)) or sin2 x =
2
1
(1 − cos (2x)).
2
Evaluate:
Z
sin5 x cos2 x dx
Solution. Using sin2 x = 1 − cos2 x one writes
Z
sin5 x cos2 x dx =
=
Z h
sin2 x cos2 x sin x dx
Z h
1 − cos2 x cos2 x sin x dx
i2
i2
Using the substitution u = cos x and du = − sin x dx one can now write
Z h
i2
2
2
1 − cos x cos x sin x dx = −
= −
=
=
Z h
1 − cos2 x cos2 x (− sin x) dx
Z h
1 − u2 u2 du
i2
i2
Z h
1 − 2u2 + u4
Z h
−u2 + 2u4 − u6 du
= −
i
−u2 du
i
u3 2u5 u7
+
−
+C
3
5
7
Thus in terms of x,
Z
sin5 x cos2 x dx = −
cos3 x 2 cos5 x cos7 x
+
−
+C .
3
5
7
Verify the following:
Z
π/2
sin2 x dx =
0
π
4
Solution. Using sin2 x = 1 − cos2 x one writes
Z
π/2
2
sin x dx =
0
Z
π/2
0
=
Z
0
π/2
1
(1 − cos (2x)) dx
2
1 1
− cos (2x) dx
2 2
π/2
1 π/2 1 sin (2x) =
x| −
2 0
2
2 0
π
1
= ( − 0 ) − ( sin π − sin 0 )
4
4
π
−0
=
4
π
.
=
4
Thus we have verified that
Z
π/2
sin2 x dx =
0
π
.
4
Evaluate:
Z
π/2
0
cos t
dt
1 + sin t
Solution. Using u-substitution with u = 1 + sin t and du = cos t dt one writes upon
determining that when t = 0 then u = 1 and when t = π/2 then u = 2
Z
0
π/2
Z π/2
cos t
1
dt =
cos t, dt
1 + sin t
1 + sin t
0
Z 2
1
=
du
1 u
= ln (u)|21
= ln 2 − ln 1
= ln 2
Thus
Z
0
π/2
cos t
dt = ln 2 .
1 + sin t
MATH 1020 WORKSHEET 7.4
Partial Fractions
Step
Step
Step
Step
1:
2:
3:
4:
Long Division if Degree of Numerator ≥ Degree of Denominator
Factor Denominator
Partial Fraction Decomposition
Integrate
x3 − x + 3
dx
x2 + x − 2
Solution. Noting that the degree of the
nator, we must first do long division.
x
2
3
x + x − 2 x
− x
3
2
− (x
+ x
− 2x )
− x2
+x
2
− (− x
− x
2x
So we can write
Z
numerator is > the degree of the denomi− 1
+ 3
+3
+ 2)
+ 1
Z
x3 − x + 3
2x + 1
dx
=
(x − 1) + 2
dx
2
x +x−2
x +x−2
Z
Given that the denominator factors as x2 + x − 2 = (x + 2) (x − 1) the Partial
Fraction Decomposition is as follows:
A
B
2x + 1
=
+
+x−2
x+2 x−1
2x + 1 = A(x − 1) + B(x + 2)
x2
One notes that if x = 1 the resulting equation is 3 = A·0+B(3) which leads to B = 1.
If one substitutes x = −2 into the above equation, the result gives −3 = A(−3)+B ·0,
or A = 1. Proceeding with our integration we have
Z
Z
x3 − x + 3
2x + 1
dx =
(x − 1) + 2
dx
2
x +x−2
x +x−2
Z
1
1
=
(x − 1) +
+
dx
x+2 x−1
x2
=
− x + ln |x + 2| + ln |x − 1| + C
2
Thus
Z
x3 − x + 3
x2
dx
=
− x + ln |x + 2| + ln |x − 1| + C .
x2 + x − 2
2
Z
x2 − 2
dx
(x + 1)(x2 + 1)
Solution. We observe that we do not need to complete Steps 1 and 2 for this problem.
We therefore start with Step 3, Partial Fraction Decomposition.
x2 − 2
A
Bx + C
=
+ 2
2
(x + 1)(x + 1)
x+1
x +1
2
2
x − 2 = A(x + 1) + (Bx + C)(x + 1)
= Ax2 + A + Bx2 + Bx + Cx + C
= (A + B)x2 + (B + C)x + (A + C)
This gives us the following system of equations to solve
1 = A+B
0 = B+C
−2 = A + C
x2 coefficients
x coefficients
constants
Using the x equation to determine that B = −C and subbing into the x2 equation
we can reduce the system to two equations with two unknowns.
1 = A−C
−2 = A + C
−1 = 2A
1
3
3
Thus A = − , B = and −C = . For our integration step we have
2
2
2
Z
Z
1 1
3 x−1
x2 − 2
dx
=
−
+
dx
2
(x + 1)(x + 1)
2 x + 1 2 x2 + 1
Z
3 x
3 1
1 1
+
−
dx
=
−
2
2 x + 1 2 x + 1 2 x2 + 1
1Z
1
3 Z 2x
3Z
1
= −
dx +
dx
−
dx
2
2
2 x+1
4 x +1
2 x +1
1
3
3
= − ln |x + 1| + ln |x2 + 1| − arctan x + C
2
4
2
Thus
Z
x2 − 2
1
3
3
2
dx
=
−
ln
|x
+
1|
+
ln
|x
+
1|
−
arctan x + C .
(x + 1)(x2 + 1)
2
4
2
MATH 1020 WORKSHEET 7.3
Trigonometric Substitution
These inverse substitutions use properties of trig functions to aid in solving integrals.
We will use substitutions based on the table below.
Expression
Substitution
√
a2 − x 2
x = a sin θ, − π2 ≤ θ ≤ π2
√
a2 + x 2
x = a tan θ, − π2 < θ < π2
√
x2 − a2 x = a sec θ, 0 ≤ θ ≤ π2 or π ≤ θ ≤
Identity
1 − sin2 θ = cos2 θ
1 + tan2 θ = sec2 θ
3π
2
sec2 θ − 1 = tan2 θ
Use the substitution provided to evaluate the following integrals:
Z √ 2
x −9
dx;
x = 3 sec θ
x4
Solution. Using inverse substitution with x = 3 sec θ, one finds that dx = 3 sec θ tan θ dθ
and that x2 − 9 = 9 sec2 θ − 9 = 9 tan2 θ. This allows one to write
Z
Z √ 2
x −9
3 tan θ
dx
=
3 sec θ tan θ dθ
4
x
81 sec4 θ
1 Z tan2 θ
dθ
=
9 Z sec3 θ
1
1
=
tan2 θ
dθ
9
sec3 θ
sin2 θ cos3 θ
1Z
dθ
=
9 Z cos2 theta 1
1
=
sin2 θ cos θ dθ
9
From here we solve by letting u = sin θ and du = cos θ dθ giving us
1Z
1Z
sin2 θ cos θ dθ =
sin2 θ cos θ, dθ
9
9
1Z 2
=
u du
9
1 u3
=
+C
9 3
1
=
sin3 θ + C
27
The last step is our answer in terms of θ. We
must now use a triangle to determine the answer in terms of x, our original variable. Thus
√
x2 − 9
1
dx =
3
x
27
√
x
θ
9 − x2
x
3
!3
+C .
√
9 − x2
Z
du
√
;
2
u 5 − u2
u=
√
5 sin θ
√
√
Solution. Using inverse substitution with u = 5 sin θ, one finds that du = 5 cos θ dθ
and that 5 − u2 = 5 − 5 sin2 θ = 5 cos2 θ. This allows one to write
√
Z
Z
du
5 cos θ
√
√
=
dθ
2
u2 5 − u 2
5 sin θ 5 cos θ
Z
dθ
=
5 sin2 θ
Z
1
=
csc2 θ dθ
5
1
=
cot2 θ + C
5
√
Using a triangle to determine the answer in
terms of u, our original variable, we find that
Z
du
1
√
=
2
2
5
u 5−u
5
θ
√
√
5 − u2
u
5 − u2
!
+C .
u
MATH 1020 WORKSHEET 7.8
Improper Integrals
∞
Improper integrals either have infinite limits of integration,
x dx, or the integrand
R 2 a1
approaches ∞ or −∞ inside the interval of integration, 0 x−1 dx, or both. Each of
these infinities must be replaced by at least one limit! In solving improper integrals
(1) replace infinities with limits, (2) integrate, (3) evaluate the integral, (4) take the
limit and (5) write a conclusion.
R
In the integrals below, determine if the improper integral converges or diverges by
first replacing each infinity with a limit.
Z
∞
(x − 1)e−x dx
0
Solution. We observe that the only infinity is in this problem is on the upper limit
of integration. Thus we write
Z
∞
1
(x − 1)e−x dx = lim
Z
t
t→∞ 0
0
(x − 1)e−x dx
t
2
= lim −xe−x t→∞
3
−t
= lim −te
t→∞
0
−0
−t
t→∞ et
= lim
−∞
which is an indeterminate form. Applying
∞
L’Hospital’s rule (see section 4.4 page 299) one finds
−t L0 H
−1
lim t = lim t
t→∞ e
t→∞ e
4
= 0
When evaluating this limit one obtains
Thus we can conclude (5)
The integral
Z
∞
(x − 1)e−x dx converges to 0 .
0
Note that step (2) integration is accomplished using integration by parts with u = x−1
and dv = e−x dx and the following calculations
Z
−x
(x − 1)e
dx =
Z
(x − 1) e−x , dx
= −(x − 1)e−x −
−x
= −(x − 1)e
+
Z
−e−x dx
Z
e−x dx
= −(x − 1)e−x − e−x + C
= −xe−x + e−x − e−x + C
= −xe−x + C
Z
8
0
√
3
1
dx
8−x
Solution. We observe that the integrand is undefined at x = 8. Thus we write
Z
0
8
√
3
1
1
dx =
8−x
lim−
t→8
Z
t
0
√
3
1
dx
8−x
t
3
lim− − (8 − x)2/3 t→8
2
0
3
3
3
2/3
2/3
= lim− − (8 − x) + (8)
t→8
2
2
3 2
4
= 0 + (2)
2
= 6
2
=
Thus we can conclude (5)
The integral
Z
0
8
√
3
1
dx converges to 6 .
8−x
Note that step (2) integration is accomplished using u-substitution letting u = 8 − x
and du = −dx resulting in
Z
√
3
Z
1
1
dx = − √
du
3
u
8−x
= −
Z
u−1/3 du
u2/3
+C
2/3
3
= − u2/3 + C
2
3
= − (8 − x)2/3 + C
2
= −