Download §8.2–Trigonometric Integrals

§8.2–Trigonometric Integrals
Mark Woodard
Furman U
Fall 2010
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§8.2–Trigonometric Integrals
Fall 2010
1 / 14
Outline
1
Powers of sine and cosine
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§8.2–Trigonometric Integrals
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Powers of sine and cosine
Essential identities
Here are the basic identities that we need:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Essential identities
Here are the basic identities that we need:
cos2 (x) + sin2 (x) = 1
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
3 / 14
Powers of sine and cosine
Essential identities
Here are the basic identities that we need:
cos2 (x) + sin2 (x) = 1
1 + cos(2x)
cos2 (x) =
2
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
3 / 14
Powers of sine and cosine
Essential identities
Here are the basic identities that we need:
cos2 (x) + sin2 (x) = 1
1 + cos(2x)
cos2 (x) =
2
1
−
cos(2x)
sin2 (x) =
2
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
3 / 14
Powers of sine and cosine
Essential identities
Here are the basic identities that we need:
cos2 (x) + sin2 (x) = 1
1 + cos(2x)
cos2 (x) =
2
1
−
cos(2x)
sin2 (x) =
2
1 + tan2 (x) = sec2 (x)
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
3 / 14
Powers of sine and cosine
Powers of sine and cosine, Case 1
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Powers of sine and cosine, Case 1
Z
Consider an integral of the form
Mark Woodard (Furman U)
cosm (x) sinn (x)dx: m or n is odd.
§8.2–Trigonometric Integrals
Fall 2010
4 / 14
Powers of sine and cosine
Powers of sine and cosine, Case 1
Z
Consider an integral of the form
cosm (x) sinn (x)dx: m or n is odd.
We use the identity sin2 (x) + cos2 (x) = 1 and u-substitution.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
4 / 14
Powers of sine and cosine
Powers of sine and cosine, Case 1
Z
Consider an integral of the form
cosm (x) sinn (x)dx: m or n is odd.
We use the identity sin2 (x) + cos2 (x) = 1 and u-substitution.
Problem
Z
Evaluate I =
sin4 (x) cos7 (x)dx.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Solution
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Solution
Write
sin4 (x) cos7 (x) = sin4 (x) cos2 (x)
3
cos(x)
= sin4 (x) 1 − sin2 (x)
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
3
cos(x).
Fall 2010
5 / 14
Powers of sine and cosine
Solution
Write
sin4 (x) cos7 (x) = sin4 (x) cos2 (x)
3
cos(x)
3
= sin4 (x) 1 − sin2 (x) cos(x).
Z
Let u = sin(x). Then du = cos(x)dx and I = u 4 (1 − u 2 )3 du.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
5 / 14
Powers of sine and cosine
Solution
Write
sin4 (x) cos7 (x) = sin4 (x) cos2 (x)
3
cos(x)
3
= sin4 (x) 1 − sin2 (x) cos(x).
Z
Let u = sin(x). Then du = cos(x)dx and I = u 4 (1 − u 2 )3 du.
Then
Z
I =
u 4 − 3u 6 + 3u 8 − u 10 du
u 5 3u 7 3u 9 u 11
−
+
−
+C
5
7
9
11
sin5 (x) 3 sin7 (x) 3 sin9 (x) sin11 (x)
=
−
+
−
+C
5
7
9
11
=
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Powers of sine and cosine, Case 2
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§8.2–Trigonometric Integrals
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Powers of sine and cosine
Powers of sine and cosine, Case 2
Z
Consider an integral of the form
Mark Woodard (Furman U)
cosm (x) sinn (x)dx: m and n even.
§8.2–Trigonometric Integrals
Fall 2010
6 / 14
Powers of sine and cosine
Powers of sine and cosine, Case 2
Z
Consider an integral of the form
cosm (x) sinn (x)dx: m and n even.
Use the the half-angle identities, repeatedly if necessary.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
6 / 14
Powers of sine and cosine
Powers of sine and cosine, Case 2
Z
Consider an integral of the form
cosm (x) sinn (x)dx: m and n even.
Use the the half-angle identities, repeatedly if necessary.
Problem
Z
Solve the integral
π/2
sin2 (x)dx.
0
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
6 / 14
Powers of sine and cosine
Solution
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§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Solution
Z
We will solve the anti-derivative problem I =
sin2 (x)dx first and
evaluate at the end.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
7 / 14
Powers of sine and cosine
Solution
Z
We will solve the anti-derivative problem I =
sin2 (x)dx first and
evaluate at the end.
Since sin2 (x) = (1 − cos(2x))/2, we have
Z
1 − cos(2x)
x
sin(2x)
I =
dx = −
+ C.
2
2
4
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
7 / 14
Powers of sine and cosine
Solution
Z
We will solve the anti-derivative problem I =
sin2 (x)dx first and
evaluate at the end.
Since sin2 (x) = (1 − cos(2x))/2, we have
Z
1 − cos(2x)
x
sin(2x)
I =
dx = −
+ C.
2
2
4
We can drop the “+C” and evaluate the resulting anti-derivative
between the limits, obtaining
Z
π/2
sin2 (x)dx =
0
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
π
.
4
Fall 2010
7 / 14
Powers of sine and cosine
Problem
Z
Evaluate I =
sin2 (x) cos2 (x)dx.
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§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Problem
Z
Evaluate I =
sin2 (x) cos2 (x)dx.
Solution
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
8 / 14
Powers of sine and cosine
Problem
Z
Evaluate I =
sin2 (x) cos2 (x)dx.
Solution
By two applications of the double-angle formulas we find
1 − cos2 (2x)
4
1 cos(4x)
= −
.
8
8
sin2 (x) cos2 (x) =
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§8.2–Trigonometric Integrals
Fall 2010
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Powers of sine and cosine
Problem
Z
Evaluate I =
sin2 (x) cos2 (x)dx.
Solution
By two applications of the double-angle formulas we find
1 − cos2 (2x)
4
1 cos(4x)
= −
.
8
8
sin2 (x) cos2 (x) =
Finally
Z I =
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1 cos(4x)
−
8
8
dx =
§8.2–Trigonometric Integrals
sin(4x)
x
−
+C
8
32
Fall 2010
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Powers of sine and cosine
Powers of tangent and secant, Case 1
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
9 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 1
Z
Consider an integral of the form
tanm (x) secn (x)dx where n, the
power of the secant, is even.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
9 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 1
Z
Consider an integral of the form
tanm (x) secn (x)dx where n, the
power of the secant, is even.
In this case, keep a sec2 (x) and convert the remaining secants to
tangents through 1 + tan2 (x) = sec2 (x).
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
9 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 1
Z
Consider an integral of the form
tanm (x) secn (x)dx where n, the
power of the secant, is even.
In this case, keep a sec2 (x) and convert the remaining secants to
tangents through 1 + tan2 (x) = sec2 (x).
Make a u-substitution: u = tan(x), du = sec2 (x)dx.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
9 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 1
Z
Consider an integral of the form
tanm (x) secn (x)dx where n, the
power of the secant, is even.
In this case, keep a sec2 (x) and convert the remaining secants to
tangents through 1 + tan2 (x) = sec2 (x).
Make a u-substitution: u = tan(x), du = sec2 (x)dx.
Problem
Z
Solve I =
π/4
tan4 (x) sec6 (x)dx.
0
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
9 / 14
Powers of sine and cosine
Solution
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
10 / 14
Powers of sine and cosine
Solution
Write
2
tan4 (x) sec6 (x) = tan4 (x) sec2 (x) sec2 (x)
2
= tan4 (x) 1 + tan2 (x) sec2 (x).
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
10 / 14
Powers of sine and cosine
Solution
Write
2
tan4 (x) sec6 (x) = tan4 (x) sec2 (x) sec2 (x)
2
= tan4 (x) 1 + tan2 (x) sec2 (x).
We make the substitution u = tan(x) and du = sec2 (x)dx; thus,
Z
1
I =
4
2 2
Z
u (1 + u ) du =
0
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1
(u 4 + 2u 6 + u 8 )du =
0
§8.2–Trigonometric Integrals
143
315
Fall 2010
10 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 2
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
11 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 2
Z
Consider an integral of the form
tanm (x) secn (x)dx where m, the
power of the tangent, is odd.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
11 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 2
Z
Consider an integral of the form
tanm (x) secn (x)dx where m, the
power of the tangent, is odd.
In this case, keep one tangent and convert the remaining tangents to
secants through 1 + tan2 (x) = sec2 (x).
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
11 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 2
Z
Consider an integral of the form
tanm (x) secn (x)dx where m, the
power of the tangent, is odd.
In this case, keep one tangent and convert the remaining tangents to
secants through 1 + tan2 (x) = sec2 (x).
Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
11 / 14
Powers of sine and cosine
Powers of tangent and secant, Case 2
Z
Consider an integral of the form
tanm (x) secn (x)dx where m, the
power of the tangent, is odd.
In this case, keep one tangent and convert the remaining tangents to
secants through 1 + tan2 (x) = sec2 (x).
Make a u-substitution: u = sec(x), du = sec(x) tan(x)dx.
Problem
Z
Solve
π/3
tan7 (x) sec5 (x)dx.
0
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
11 / 14
Powers of sine and cosine
Solution
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§8.2–Trigonometric Integrals
Fall 2010
12 / 14
Powers of sine and cosine
Solution
Write
3
tan7 (x) sec5 (x) = tan(x) tan2 (x) sec5 (x)
= tan(x) sec2 (x) − 1)3 sec5 (x)
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
12 / 14
Powers of sine and cosine
Solution
Write
3
tan7 (x) sec5 (x) = tan(x) tan2 (x) sec5 (x)
= tan(x) sec2 (x) − 1)3 sec5 (x)
If u = sec(x), then du = tan(x) sec(x)dx and
2
Z
(u 2 − 1)3 u 4 du
I =
1
2
Z
(u 10 − 3u 8 + 3u 6 − u 4 )du
=
1
4779
=
40
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
12 / 14
Powers of sine and cosine
A note on powers of tangents and secants
Our analysis is not exhaustive:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
13 / 14
Powers of sine and cosine
A note on powers of tangents and secants
Our analysis is not exhaustive:
We do not have any direct methods for integrating powers of tangent
alone and powers of secant alone, for example,
Z
Z
8
tan (x)dx and
sec4 (x)dx.
These integrals can be solved, but we will not address these cases by
direct methods.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
13 / 14
Powers of sine and cosine
A note on powers of tangents and secants
Our analysis is not exhaustive:
We do not have any direct methods for integrating powers of tangent
alone and powers of secant alone, for example,
Z
Z
8
tan (x)dx and
sec4 (x)dx.
These integrals can be solved, but we will not address these cases by
direct methods.
We do not have any methods for integrating an even power of
tangent times an odd power of secant, for example,
Z
tan4 (x) sec5 (x)dx.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
13 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
Show that
R
sec(x)dx = ln | sec(x) + tan(x)| + C
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
R
Evaluate tan2 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
R
Evaluate tan2 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
R
Evaluate tan2 (x)dx Hint: use a nice trig identity.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
R
Evaluate tan2 (x)dx Hint: use a nice trig identity.
R
Evaluate tan3 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
R
Evaluate tan2 (x)dx Hint: use a nice trig identity.
R
Evaluate tan3 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
R
Show that sec(x)dx = ln | sec(x) + tan(x)| + C
R
Evaluate sec2 (x)dx Hint: this is easy!
R
Evaluate tan2 (x)dx Hint: use a nice trig identity.
R
Evaluate tan3 (x)dx Hint: save out a tangent and then use a trig
identity.
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
Show that
R
Evaluate
R
Evaluate
R
Evaluate
identity.
R
Evaluate
R
sec(x)dx = ln | sec(x) + tan(x)| + C
sec2 (x)dx Hint: this is easy!
tan2 (x)dx Hint: use a nice trig identity.
tan3 (x)dx Hint: save out a tangent and then use a trig
sec3 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
Show that
R
Evaluate
R
Evaluate
R
Evaluate
identity.
R
Evaluate
R
sec(x)dx = ln | sec(x) + tan(x)| + C
sec2 (x)dx Hint: this is easy!
tan2 (x)dx Hint: use a nice trig identity.
tan3 (x)dx Hint: save out a tangent and then use a trig
sec3 (x)dx Hint:
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14
Powers of sine and cosine
Problem (Some ad hoc problems)
Show that
R
tan(x)dx = ln | sec(x)| + C
Show that
R
Evaluate
R
Evaluate
R
Evaluate
identity.
R
Evaluate
R
sec(x)dx = ln | sec(x) + tan(x)| + C
sec2 (x)dx Hint: this is easy!
tan2 (x)dx Hint: use a nice trig identity.
tan3 (x)dx Hint: save out a tangent and then use a trig
sec3 (x)dx Hint: try parts with u = sec x. (see page 500.)
Mark Woodard (Furman U)
§8.2–Trigonometric Integrals
Fall 2010
14 / 14