Applied Calculus I
Lecture 29
Integrals of trigonometric functions
We shall continue learning substitutions by considering integrals involving
trigonometric functions.
Integrals of trigonometric functions
We shall continue learning substitutions by considering integrals involving
trigonometric functions.
Z
Find sin (x2 − 3)xdx
Integrals of trigonometric functions
We shall continue learning substitutions by considering integrals involving
trigonometric functions.
Z
Find sin (x2 − 3)xdx
The expression that seems to complicate things here is x2 − 3. So, let us
make the substitution u = x2 − 3. Then du = 2xdx, whic implies
1
xdx = 2 du.
Integrals of trigonometric functions
We shall continue learning substitutions by considering integrals involving
trigonometric functions.
Z
Find sin (x2 − 3)xdx
The expression that seems to complicate things here is x2 − 3. So, let us
make the substitution u = x2 − 3. Then du = 2xdx, whic implies
1
xdx = 2 du.
Then our integral becomes
Z
1
1
1
sin udu = − cos u + C = − cos (x2 − 3) + C
2
2
2
Examples
Z
Find
tan xdx
Examples
Z
Find
tan xdx
Z
Z
sin x
Notice that tan xdx =
dx. The problematic part is the
cos x
denominator. So, let u = cos x. Then du = − sin xdx.
Examples
Z
Find
tan xdx
Z
Z
sin x
Notice that tan xdx =
dx. The problematic part is the
cos x
denominator. So, let u = cos x. Then du = − sin xdx.
Now we see that
Z our integral becomes
−du
= − ln |u| + C = − ln | cos x| + C
u
Examples
Z
Find
tan xdx
Z
Z
sin x
Notice that tan xdx =
dx. The problematic part is the
cos x
denominator. So, let u = cos x. Then du = − sin xdx.
Now we see that
Z our integral becomes
−du
= − ln |u| + C = − ln | cos x| + C
u
Z
Find cot xdx
Examples
Z
Find
tan xdx
Z
Z
sin x
Notice that tan xdx =
dx. The problematic part is the
cos x
denominator. So, let u = cos x. Then du = − sin xdx.
Now we see that
Z our integral becomes
−du
= − ln |u| + C = − ln | cos x| + C
u
Z
Find cot xdx
Z
Z
cos x
Notice that cot xdx =
dx. The problematic part is the
sin x
denominator. So, let u = sin x. Then du = cos xdx.
Examples
Z
Find
tan xdx
Z
Z
sin x
Notice that tan xdx =
dx. The problematic part is the
cos x
denominator. So, let u = cos x. Then du = − sin xdx.
Now we see that
Z our integral becomes
−du
= − ln |u| + C = − ln | cos x| + C
u
Z
Find cot xdx
Z
Z
cos x
Notice that cot xdx =
dx. The problematic part is the
sin x
denominator. So, let u = sin x. Then du = cos xdx.
Now we see that our
Z integral becomes
du
= ln |u| + C = ln | sin x| + C
u
Z
Find
Examples
dx
x ln x
Z
Find
Examples
dx
x ln x
Again, the part that complicates things is the denominator. But if we try
the substitution u = x ln x we see that du = (ln x + 1)dx, and then we
can’t express everything in terms of u.
Z
Find
Examples
dx
x ln x
Again, the part that complicates things is the denominator. But if we try
the substitution u = x ln x we see that du = (ln x + 1)dx, and then we
can’t express everything in terms of u.
The idea is to make a substitution for the part of your integrand whose
derivative gives you the rest of the integrand (perhaps multiplied by a
constant).
Z
Find
Examples
dx
x ln x
Again, the part that complicates things is the denominator. But if we try
the substitution u = x ln x we see that du = (ln x + 1)dx, and then we
can’t express everything in terms of u.
The idea is to make a substitution for the part of your integrand whose
derivative gives you the rest of the integrand (perhaps multiplied by a
constant).
0
In this case, we notice that (ln x) =
du = x1 dx, and our integral becomes
1
.
x
So, let us try u = ln x. Then
Z
Find
Examples
dx
x ln x
Again, the part that complicates things is the denominator. But if we try
the substitution u = x ln x we see that du = (ln x + 1)dx, and then we
can’t express everything in terms of u.
The idea is to make a substitution for the part of your integrand whose
derivative gives you the rest of the integrand (perhaps multiplied by a
constant).
0
1
.
x
So, let us try u = ln x. Then
In this case, we notice that (ln x) =
du = x1 dx, and our integral becomes
Z
du
= ln |u| + C = ln | ln x| + C
u
Examples
Z
Find
3x2 +1
x8
dx
Examples
Z
Find
3x2 +1
x8
dx
Let u = 3x2 + 1, then du = 6xdx. Hence,
Z
Z
u
3x2 +1
1
8
1 8
3x2 +1
u
x8
dx =
8 du = ·
+C =
+C
6
6 ln 8
6 ln 8
Examples
Z
Find
3x2 +1
x8
dx
Let u = 3x2 + 1, then du = 6xdx. Hence,
Z
Z
u
3x2 +1
1
8
1 8
3x2 +1
u
x8
dx =
8 du = ·
+C =
+C
6
6 ln 8
6 ln 8
Z
Find
7
sin x cos xdx
Examples
Z
Find
3x2 +1
x8
dx
Let u = 3x2 + 1, then du = 6xdx. Hence,
Z
Z
u
3x2 +1
1
8
1 8
3x2 +1
u
x8
dx =
8 du = ·
+C =
+C
6
6 ln 8
6 ln 8
Z
Find
7
sin x cos xdx
Let u = sin x, then du = cos xdx, yielding
Z
Z
1 8
1 8
7
7
sin x cos xdx = u du = u + C = sin x + C
8
8
Z
Find
cos3 x
dx
sin x + 1
Examples
Examples
cos3 x
Find
dx
sin x + 1
Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx
becomes du, the denominator becomes u, but we still have the factor
cos2 x that we need to express in terms of u.
Z
Examples
cos3 x
Find
dx
sin x + 1
Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx
becomes du, the denominator becomes u, but we still have the factor
cos2 x that we need to express in terms of u.
Z
Recall that cos2 x + sin2 x = 1, so cos2 x = 1 − sin2 x. Also,
sin2 x = (sin x + 1 − 1)2 = (u − 1)2 = u2 − 2u + 1. Therefore,
cos2 x = 2u − u2 .
Examples
cos3 x
Find
dx
sin x + 1
Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx
becomes du, the denominator becomes u, but we still have the factor
cos2 x that we need to express in terms of u.
Z
Recall that cos2 x + sin2 x = 1, so cos2 x = 1 − sin2 x. Also,
sin2 x = (sin x + 1 − 1)2 = (u − 1)2 = u2 − 2u + 1. Therefore,
cos2 x = 2u − u2 .
Thus, we obtain
Z
Z
Z
2
3
2u − u
cos x
1 2
dx =
du = (2 − u)du = 2u − u + C =
sin x + 1
u
2
1
2
= 2(sin x + 1) − (sin x + 1) + C
2
Examples
cos3 x
Find
dx
sin x + 1
Proceeding as usual, let u = sin x + 1, then du = cos xdx. So, cos xdx
becomes du, the denominator becomes u, but we still have the factor
cos2 x that we need to express in terms of u.
Z
Recall that cos2 x + sin2 x = 1, so cos2 x = 1 − sin2 x. Also,
sin2 x = (sin x + 1 − 1)2 = (u − 1)2 = u2 − 2u + 1. Therefore,
cos2 x = 2u − u2 .
Thus, we obtain
Z
Z
Z
2
3
2u − u
cos x
1 2
dx =
du = (2 − u)du = 2u − u + C =
sin x + 1
u
2
1
2
= 2(sin x + 1) − (sin x + 1) + C
2
In the hindsight, we could do
Z
Z
Z
2
3
cos x
(1 − sin x) cos x
dx =
dx = (1 − sin x) cos xdx
sin x + 1
1 + sin x
and then integrate.
Example
An epidemic is growing in a region according to the rate
100t
0
N (t) = 2
,
t +2
where N (t) is the number of people infected after t days. Find a formula
for the number of people infected after t dyas, given that 37 people were
infected at t = 0.
Example
An epidemic is growing in a region according to the rate
100t
0
N (t) = 2
,
t +2
where N (t) is the number of people infected after t days. Find a formula
for the number of people infected after t dyas, given that 37 people were
infected at t = 0.
So, we need a formula for N (t) given that N (0) = 37. Clearly, N (t) is
100t
some antiderivative of 2
. So, let’s find the whole family.
t +2
Example
An epidemic is growing in a region according to the rate
100t
0
N (t) = 2
,
t +2
where N (t) is the number of people infected after t days. Find a formula
for the number of people infected after t dyas, given that 37 people were
infected at t = 0.
So, we need a formula for N (t) given that N (0) = 37. Clearly, N (t) is
100t
some antiderivative of 2
. So, let’s find the whole family.
t +2
Z
u=t2 +2,du=2tdt Z
100t
50
z}|{
2
dt
=
du
=
50
ln
|u|
+
C
=
50
ln
|t
+ 2| + C
2
t +2
u
Example
An epidemic is growing in a region according to the rate
100t
0
N (t) = 2
,
t +2
where N (t) is the number of people infected after t days. Find a formula
for the number of people infected after t dyas, given that 37 people were
infected at t = 0.
So, we need a formula for N (t) given that N (0) = 37. Clearly, N (t) is
100t
some antiderivative of 2
. So, let’s find the whole family.
t +2
Z
u=t2 +2,du=2tdt Z
100t
50
z}|{
2
dt
=
du
=
50
ln
|u|
+
C
=
50
ln
|t
+ 2| + C
2
t +2
u
So, N (t) = 50 ln (t2 + 2) + C for some specific value of C. To find this
value, we need to use the fact that N (0) = 37.
Example
An epidemic is growing in a region according to the rate
100t
0
N (t) = 2
,
t +2
where N (t) is the number of people infected after t days. Find a formula
for the number of people infected after t dyas, given that 37 people were
infected at t = 0.
So, we need a formula for N (t) given that N (0) = 37. Clearly, N (t) is
100t
some antiderivative of 2
. So, let’s find the whole family.
t +2
Z
u=t2 +2,du=2tdt Z
100t
50
z}|{
2
dt
=
du
=
50
ln
|u|
+
C
=
50
ln
|t
+ 2| + C
2
t +2
u
So, N (t) = 50 ln (t2 + 2) + C for some specific value of C. To find this
value, we need to use the fact that N (0) = 37.
Plugging t = 0 into the formula for N (t) we get 50 ln 2 + C = 37 that is,
C = 37 − 50 ln 2. Therefore, N (t) = 50 ln (t2 + 2) + 37 − 50 ln 2.
2
This can be simplified to N (t) = 50 ln t2 + 1 + 37