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Trigonometric Integrals 1
Lecture Notes
page 1
Sample Problems
Compute each of the following integrals. Assume that a and b are positive numbers.
1.
Z
2.
Z
sin x dx
cos 5x dx
3.
Z
4.
Z
csc2 x
5.
Z
tan x dx
6.
Z
7.
Z
cos x sin4 x dx
dx
cot x dx
sec x dx
8.
Z
csc x dx
9.
Z
sin2 x dx
10.
Z
sin3 x dx
11.
Z
sin x dx
12.
Z
sin5 x dx
13.
Z
1
dx
2
a + b2 x2
14.
Z
15.
Z
p
sec ( x)
p
dx
x
16.
Z=3
p
1 + cos 2x dx
Z=2
p
1
0
17.
4
cos x dx
0
p
1
a2
x2
18.
dx
Z
tan3 x dx
19.
Z
sin 7x cos 3x dx
20.
Z
sin 10x sin 4x dx
15.
Z=6
Practice Problems
1.
2.
3.
4.
5.
6.
7.
Z
Z
Z
Z
Z
Z
Z
cos 3x dx
sin 4x
8.
5
dx
sec tan d
sec2
d
x tan x2
cot (2x
cos2 x
9.
10.
11.
dx
) dx
dx
c copyright Hidegkuti, Powell, 2012
12.
13.
14.
Z
Z
Z
Z
Z
Z
Z
cos2 (2x) dx
16.
sin 2a cos 8a da
17.
Z
cos b cos 11b db
18.
Z
sin 6 sin 14 d
19.
Z
cos 11m sin 3m dm
sin x cos5 x dx
tan2 x dx
cos 6x dx
Z
cos5 x dx
sin3 x cos5 x dx
1
0
cos3 x dx
cos4 x dx
p
Last revised: December 8, 2013
Trigonometric Integrals 1
Lecture Notes
page 2
Sample Problems - Answers
1.)
cos x + C
1
sin 5x + C
5
2.)
6.) ln jsin xj + C
3.)
7.) ln jsec x + tan xj + C
10.)
1
cos3 x
3
cos x + C
11.)
13.)
1
tan
ab
b
x +C
a
14.) sin
1
p
17.) 2 2 2
1
sin5 x + C
5
18.)
8.)
4.)
cot x + C
ln jcsc x + cot xj + C
1
1
3
sin 2x +
sin 4x + x + C
4
32
8
1
12.)
ln jcos xj + C = ln jsec xj + C
1
x
2
9.)
cos x +
1
sin 2x + C
4
2
cos3 x
3
p
p
15.) 2 ln jsec ( x) + tan ( x)j + C
x
+C
a
1
sec2 x + ln jcos xj + C
2
5.)
19.)
1
cos 10x
20
1
cos 4x + C
8
20.)
1
cos5 x + C
5
p
6
16.)
2
1
sin 6x
12
1
sin 14x + C
28
Practice Problems - Answers
1.)
1
sin 3x + C
3
2.)
6.)
1
ln jsin (2x
2
)j + C
10.)
1
cos 4x
4
5
3
1
1
x + sin 2x +
sin 4x + C
8
4
32
1
1
cos6 x + cos8 x + C
6
8
17.)
1
1
sin 10b +
sin 12b + C
20
24
3.) sec + C
1
1
x + sin 2x + C
2
4
7.)
13.)
+C
c copyright Hidegkuti, Powell, 2012
18.)
1
1
x + sin 4x + C
2
8
5.)
1
ln sec x2
2
9.) sin x
+C
1
sin3 x + C
3
2
1
sin3 x + sin x + C
12.)
cos6 x + C
3
6
p
2
1
1
x + tan x + C
15.)
16.)
cos 6a
cos 10a + C
3
12
20
11.)
14.)
8.)
4.) tan + C
1
sin5 x
5
1
sin 8
16
1
sin 20 + C
40
19.)
1
cos 8m
16
1
cos 14m + C
28
Last revised: December 8, 2013
Trigonometric Integrals 1
Lecture Notes
page 3
Sample Problems - Solutions
1.
Z
sin x dx
Solution: This is a basic integral we know from dierentiating
basic trigonometric functions. Since
Z
d
d
cos x = sin x, clearly
( cos x) = sin x and so
sin x dx = cos x + C .
dx
dx
Z
2.
cos 5x dx
d
Solution: We know that
cos x = sin x + C. We will use substitution. Let u = 5x and then du = 5dx
dx
du
and so
= dx.
5
Z
Z
Z
du
1
1
cos 5x dx = cos u
=
cos u du = sin 5x + C
5
5
5
Note: Once we
Z have enough practice, there is no need to perform this substitution in writing. We can just
1
simply write
cos 5x dx = sin 5x + C.
5
Z
3.
cos x sin4 x dx
Solution: Let u = sin x. Then du = cos xdx.
Z
Z
Z
1
1
cos x sin4 x dx = sin4 x (cos xdx) = u4 u = u5 + C = sin5 x + C
5
5
4.
Z
csc2 x dx
d
cot x =
dx
Z
Z
2
csc x dx =
Solution: We need to remember that
5.
Z
csc2 x.
csc2 x dx =
tan x dx
Solution: Let u = cos x. Then du = sin x dx.
Z
Z
Z
Z
1
1
sin x
tan x dx =
dx =
(sin xdu) =
( du) =
cos x
u
u
= ln (cos x)
6.
Z
cot x + C
1
Z
1
du =
u
ln juj + C =
ln jcos xj + C
+ C = ln jsec xj + C
cot x dx
Solution: Let u = sin x. Then du = cos x dx.
Z
Z
Z
Z
cos x
1
1
cot x dx =
dx =
(cos xdu) =
du = ln juj + C = ln jsin xj + C
sin x
u
u
c copyright Hidegkuti, Powell, 2012
Last revised: December 8, 2013
Trigonometric Integrals 1
Lecture Notes
7.
Z
sec x dx
Z
Solution:
sec x dx =
Z
sec2 x
Z
sec2 x + sec x tan x
dx =
sec x + tan x
Z
1
u
dx.
2
sec x + sec x tan x dx =
Z
csc x dx =
Z
Z
1
du = ln juj + C = ln jsec x + tan xj + C
u
csc2 x + csc x cot x
dx =
csc x + cot x
=
Z
csc2 x + csc x cot x
dx
csc x + cot x
d
csc x =
dx
csc x cot x dx.
Recall that
csc2 x
u = csc x + cot x. Then du =
Z
Z
csc x + cot x
dx =
csc x + cot x
csc x
From here we will use substitution.
9.
1
u
csc2 x + csc x cot x dx =
ln jcsc x + cot xj + C
1
(1
2
cos 2x)
1 dx
Z
sin2 x =
Z
2
sin x dx =
=
10.
csc x cot x and
d
cot x =
dx
csc2 x.
Z
Z
ln juj + C
1
( du) =
u
1
du =
u
Let
sin2 x dx
Solution: Recall the double angle formula for cosine, cos 2x = 1
Z
Let u =
csc x dx
Solution:
Z
sec2 x + sec x tan x
dx
sec x + tan x
d
d
sec x = sec x tan x and
tan x = sec2 x.
dx
dx
Recall that
sec x + tan x. Then du = sec x tan x +
Z
Z
sec x + tan x
sec x
dx =
sec x + tan x
From here we will use substitution.
8.
page 4
Z
1
(1
2
1
cos 2x) dx =
2
1
sin 2x + C
4
1
x
2
Z
2 sin2 x. We solve this for sin2 x
cos 2x dx
=
1
2
x + C1
1
sin 2x + C2
2
sin3 x dx
Solution:
Z
3
sin x dx =
Z
2
sin x sin x dx =
Let u = cos x: Then du = sin xdx
Z
Z
Z
3
2
sin x dx =
sin x 1 cos x dx =
=
1 3
u
3
u+C =
c copyright Hidegkuti, Powell, 2012
1
cos3 x
3
1
2
Z
sin x 1
cos x (sin xdx) =
cos2 x dx
Z
1
u
2
( du) =
Z
u2
1 du
cos x + C
Last revised: December 8, 2013
Trigonometric Integrals 1
Lecture Notes
11.
Z
page 5
sin4 x dx
Solution: We use the double angle formula for cosine to express sin2 x.
2 sin2 x
cos 2x = 1
Z
4
sin x dx =
Z
2
sin x
2
dx =
Z
=)
2
1
(1
2
cos 2x)
sin2 x =
1
dx =
4
Z
1
(1
2
(1
cos 2x)
1
cos 2x) dx =
4
2
We use the double angle formula for cosine again to express cos2 2x.
Z
1
sin x dx =
4
Z
=
4
=
12.
Z
cos 4x = 2 cos2 2x
1
Z
2
1
2
=)
cos2 2x =
sin 2x +
1
8
1
4
1
2 cos 2x + cos2 2x dx
1
(cos 4x + 1)
2
Z
1
1 2 cos 2x + cos 2x dx =
1
4
Z
1 1
1
1
cos 2x + cos 4x +
dx =
4 2
8
8
1
2
Z
3
sin 4x + x + C =
8
1
(cos 4x + 1)
2
1
1
3
cos 2x + cos 4x +
2
8
8
2 cos 2x +
dx
dx
1
1
3
sin 2x +
sin 4x + x + C
4
32
8
sin5 x dx
Solution: This method works with odd powers of sin x or cos x. We will separate one factor of sin x from the
rest which will be expressed in terms of cos x.
Z
Z
Z
Z
Z
2
2
5
4
4
2
sin x dx =
sin x sin x dx = sin x sin x dx = sin x sin x dx = sin x 1 cos2 x dx
Z
=
sin x 1 2 cos2 x + cos4 x dx
We proceed with substitution. Let u = cos x: Then du = sin xdx.
Z
Z
Z
sin5 x dx =
sin x 1 2 cos2 x + cos4 x dx =
1 2 cos2 x + cos4 x (sin xdx)
Z
Z
2
1 5
=
1 2u2 + u4 ( du) =
1 + 2u2 u4 du = u + u3
u +C
3
5
1
2
=
cos x + cos3 x
cos5 x + C
3
5
13.
Z
a2
1
dx
+ b2 x2
Z
1
dx = tan
+1
a2 x2 = b2 u2 . This would be convenient because then
Solution:
The basic integral here is
x2
1x
+ C.
We need a substitution under which
1
1
1
1
= 2 2
= 2
2
2
2
+b
b u +b
b u +1
a2 x2
c copyright Hidegkuti, Powell, 2012
Last revised: December 8, 2013
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