Sample Problems - aceh.b-cdn.net
Lecture Notes
Trigonometric Integrals 1
page 1
Sample Problems
Compute each of the following integrals. Assume that a and b are positive numbers.
Z 1. sin x dx
Z 8. csc x dx
Z 15.
p
secp(
x) dx
x
Z 2. cos 5x dx
Z 3. cos x sin4 x dx
Z 9. sin2 x dx
Z 10. sin3 x dx
Z=3 p
16.
1 + cos 2x dx
0
Z 4. csc2 x dx
Z 11. sin4 x dx
Z=2 p
17.
1
0
cos x dx
Z 5. tan x dx
Z 12. sin5 x dx
Z 18. tan3 x dx
Z 6. cot x dx
Z 1
13. a2 + b2x2 dx
Z 19. sin 7x cos 3x dx
Z 7. sec x dx
Z 14. p 1 dx
a2 x2
Z 20. sin 10x sin 4x dx
Z 1. cos 3x dx
Z
2. sin 4x
dx
5
Z
3. sec tan d
Z 4. sec2 d
Z 5. x tan x2 dx
Z 6. cot (2x ) dx
Z 7. cos2 x dx
Practice Problems
Z 8. cos2 (2x) dx
Z 9. cos3 x dx
Z 10. cos4 x dx
Z 11. cos5 x dx
Z 12. sin x cos5 x dx
Z 13. sin3 x cos5 x dx
Z 14. tan2 x dx
c copyright Hidegkuti, Powell, 2012
Z=6 p
15.
1
0
cos 6x dx
Z 16. sin 2a cos 8a da
Z 17. cos b cos 11b db
Z 18. sin 6 sin 14 d
Z 19. cos 11m sin 3m dm
Last revised: December 8, 2013
Lecture Notes
Trigonometric Integrals 1
page 2
Sample Problems - Answers
1 1.) cos x + C 2.) 5 sin 5x + C
3.) 1 sin5 x + C 4.) cot x + C 5.) ln jcos xj + C = ln jsec xj + C 5
6.) ln jsin xj + C
7.) ln jsec x + tan xj + C
8.) ln jcsc x + cot xj + C
11 9.) 2 x 4 sin 2x + C
10.) 1 cos3 x cos x + C 3
13.)
1 tan 1
b x
+C
ab
a
11.) 1 sin 2x + 1 sin 4x + 3 x + C
4
32
8
12.) cos x + 2 cos3 x 1 cos5 x + C
3
5
14.) sin 1 x + C a
p
p
15.) 2 ln jsec ( x) + tan ( x)j + C
p 6
16.) 2
p 17.) 2 2 2
18.) 1 sec2 x + ln jcos xj + C
2
1
1
1
1
19.) cos 10x cos 4x + C 20.) sin 6x sin 14x + C
20
8
12
28
Practice Problems - Answers
1
1
1.) sin 3x + C 2.) cos 4x
+C
3
4
5
3.) sec + C
4.) tan + C
5.)
1 ln
sec
x2
+C
2
1 6.) ln jsin (2x )j + C
2
11 7.) x + sin 2x + C
24
11 8.) x + sin 4x + C
28
9.) sin x 1 sin3 x + C 3
31
1
10.) x + sin 2x + sin 4x + C
11.) 1 sin5 x 2 sin3 x + sin x + C
84
32
5
3
12.) 1 cos6 x + C 6
13.) 1 cos6 x + 1 cos8 x + C 14.) x + tan x + C
6
8
p 2
15.) 3
1
1
16.) cos 6a
cos 10a + C
12
20
1
1
17.) 20 sin 10b + 24 sin 12b + C
1
1
18.) 16 sin 8 40 sin 20 + C
1
1
19.) 16 cos 8m 28 cos 14m + C
c copyright Hidegkuti, Powell, 2012
Last revised: December 8, 2013
Lecture Notes
Trigonometric Integrals 1
page 3
Sample Problems - Solutions
Z 1. sin x dx
Solution: This is a basic integral we know from di?ereZntiating basic trigonometric functions. Since
d
d
dx cos x = sin x, clearly dx ( cos x) = sin x and so sin x dx = cos x + C .
Z
2. cos 5x dx
d
Solution: We know that dx cos x = sin x + C. We will use substitution. Let u = 5x and then du = 5dx
du
and so 5 = dx.
Z
Z
du
1Z
1
cos 5x dx = cos u
= cos u du = sin 5x + C
55
5
Note: Once wZ e have enough practice, there is no need to perform this substitution in writing. simply write cos 5x dx = 1 sin 5x + C.
5 Z
3. cos x sin4 x dx
We can just
Solution: Let u = sin x. Then du = cos xdx.
Z
Z
Z
cos x sin4 x dx = sin4 x (cos xdx) = u4 u = 1 u5 + C = 1 sin5 x + C
5
5
Z 4. csc2 x dx
Solution: We need to remember that d cot x = dx
Z
Z
csc2 x dx =
csc2 x. csc2 x dx =
cot x + C
Z 5. tan x dx
Solution: Let u = cos x. Then du = sin x dx.
Z
Z
Z
Z
tan x dx = sin x dx = 1 (sin xdu) = 1 ( du) =
cos x
u
u
= ln (cos x) 1 + C = ln jsec xj + C
Z 1 du = u
ln juj + C =
ln jcos xj + C
Z 6. cot x dx
Solution: Let u = sin x. Then du = cos x dx.
Z
Z
Z
Z
cot x dx =
cos x dx = sin x
1 (cos xdu) = u
1 u
du = ln juj + C =
ln jsin xj + C
c copyright Hidegkuti, Powell, 2012
Last revised: December 8, 2013
Lecture Notes
Trigonometric Integrals 1
page 4
Z 7. sec x dx
Z
Z
sec x + tan x
Z sec2 x + sec x tan x
Solution: sec x dx = sec x
dx =
dx
sec x + tan x
sec x + tan x
From here we will use substitution.
Recall that
d sec x = sec x tan x and
d tan x = sec2 x.
dx
dx
sec x + tan x. Then du = sec x tan x + sec2 x dx.
Let u =
Z sec2 x + sec x tan x
Z 1
dx =
Z
sec2 x + sec x tan x dx =
1 du = ln juj + C = ln jsec x + tan xj + C
sec x + tan x
u
u
Z 8. csc x dx
Z
Z
csc x + cot x
Z csc2 x + csc x cot x
Solution: csc x dx = csc x
dx =
dx
csc x + cot x
csc x + cot x
From here we will use substitution.
d Recall that csc x =
d csc x cot x and cot x =
csc2 x.
Let
dx
dx
u = csc x + cot x. Then du = csc2 x csc x cot x dx.
Z csc2 x + csc x cot x
Z 1
dx =
Z
csc2 x + csc x cot x dx =
1 ( du) =
Z 1 du =
ln juj + C
csc x + cot x
u
u
u
= ln jcsc x + cot xj + C
Z 9. sin2 x dx
Solution: Recall the double angle formula for cosine, cos 2x = 1 2 sin2 x. We solve this for sin2 x sin2 x = 1 (1 cos 2x) 2
Z
Z
Z
sin2 x dx =
1
1
(1 cos 2x) dx =
1 dx
2
2
11 = x sin 2x + C
24
Z 10. sin3 x dx
Z
1
cos 2x dx
= 2
x + C1
1 2 sin 2x + C2
Solution:
Z
Z
Z
sin3 x dx = sin x sin2 x dx = sin x 1 cos2 x dx
Let u = cos x: Then du = sin xdx
Z
Z
Z
Z
Z
sin3 x dx = sin x 1 cos2 x dx = 1 cos2 x (sin xdx) = 1 u2 ( du) = u2 1 du
= 1 u3 u + C = 1 cos3 x cos x + C
3
3
c copyright Hidegkuti, Powell, 2012
Last revised: December 8, 2013
Lecture Notes
Trigonometric Integrals 1
page 5
Z 11. sin4 x dx
Solution: We use the double angle formula for cosine to express sin2 x.
cos 2x = 1 2 sin2 x =) sin2 x = 1 (1 cos 2x) 2
Z
Z
Z
sin4 x dx = sin2 x 2 dx =
1 (1
2
Z 1
cos 2x) dx = (1
Z cos 2x)2 dx = 1
1
2 cos 2x + cos2 2x dx
2
4
4
We use the double angle formula for cosine again to express cos2 2x.
cos 4x = 2 cos2 2x 1
=)
cos2 2x = 1 (cos 4x + 1) 2
Z
Z
Z
sin4 x dx = 1
1
2 cos 2x + cos2 2x
1 dx =
1
Z4
4Z
11
1
1
=
cos 2x + cos 4x + dx =
42
8
8
1 2 cos 2x + (cos 4x + 1) dx
2
1 cos 2x + 1 cos 4x + 3 dx
2
8
8
= 1 1 sin 2x + 1 1 sin 4x + 3 x + C = 1 sin 2x + 1 sin 4x + 3 x + C
22
84
8
4
32
8
Z 12. 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
sin5 x dx = sin x sin4 x dx = sin x sin4 x dx = sin x sin2 x 2 dx = sin x 1 cos2 x 2 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
=
1 2u2 + u4 ( du) =
1 + 2u2 u4 du = u + 2 u3 1 u5 + C 35
=
cos x + 2 cos3 x 1 cos5 x + C
3
5
Z 1
13. a2 + b2x2 dx
Z
Solution: The basic integral here is
1 x2 + 1
dx
=
tan
1 x + C.
a2x2 = b2u2. This would be convenient because then
We need a substitution under which
1
1
11
a2x2 + b2 = b2u2 + b2 = b2 u2 + 1
c copyright Hidegkuti, Powell, 2012
Last revised: December 8, 2013
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