Download Integration using trig identities

Integration
TRIGONOMETRIC IDENTITIES
Graham S McDonald
and Silvia C Dalla
A self-contained Tutorial Module for practising
integration of expressions involving products of
trigonometric functions such as sin nx sin mx
● Table of contents
● Begin Tutorial
c 2004 [email protected]
Table of contents
1.
2.
3.
4.
5.
Theory
Exercises
Answers
Standard integrals
Tips
Full worked solutions
Section 1: Theory
3
1. Theory
Integrals of the form
Z
sin nx sin mx ,
and similar ones with products like sin nx cos mx and cos nx cos mx,
can be solved by making use of the following trigonometric identities:
1
sin A sin B = − [cos (A + B) − cos (A − B)]
2
1
sin A cos B =
[sin (A + B) + sin (A − B)]
2
1
cos A cos B =
[cos (A + B) + cos (A − B)]
2
Using these identities, such products are expressed as a sum of
trigonometric functions
This sum is generally more straightforward to integrate
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Section 2: Exercises
4
2. Exercises
Click on EXERCISE links for full worked solutions (9 exercises in
total).
Perform the following integrations:
Exercise 1.
Z
cos 3x cos 2x dx
Exercise 2.
Z
sin 5x cos 3x dx
Exercise 3.
Z
sin 6x sin 4x dx
● Theory ● Standard integrals ● Answers ● Tips
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Section 2: Exercises
Exercise 4.
Z
cos 2ωt sin ωt dt, where ω is a constant
Exercise 5.
Z
cos 4ωt cos 2ωt dt , where ω is a constant
Exercise 6.
Z
sin2 x dx
Exercise 7.
Z
sin2 ωt dt , where ω is a constant
Exercise 8.
Z
cos2 t dt
● Theory ● Standard integrals ● Answers ● Tips
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5
Section 2: Exercises
6
Exercise 9.
Z
cos2 kx dx, where k is a constant
● Theory ● Standard integrals ● Answers ● Tips
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Section 3: Answers
7
3. Answers
1.
1
10
2.
1
cos 8x − 14 cos 2x + C,
− 16
1
sin 10x + 14 sin 2x + C,
− 20
1
1
cos 3ωt + 2ω
cos ωt + C,
− 6ω
1
1
12ω sin 6ωt + 4ω sin 2ωt + C,
− 41 sin 2x + 12 x + C,
1
− 4ω
sin 2ωt + 12 t + C,
1
1
4 sin 2t + 2 t + C,
1
1
4k sin 2kx + 2 x + C.
3.
4.
5.
6.
7.
8.
9.
sin 5x +
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1
2
sin x + C,
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Section 4: Standard integrals
8
4. Standard integrals
f (x)
n
x
1
x
x
e
sin x
cos x
tan x
cosec x
sec x
sec2 x
cot x
sin2 x
cos2 x
R
f (x)dx
xn+1
n+1
(n 6= −1)
ln |x|
ex
− cos x
sin x
− ln
|cos x|
ln tan x2 ln |sec x + tan x|
tan x
ln |sin x|
x
sin 2x
2 −
4
x
sin 2x
2 +
4
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f (x)
n
0
[g (x)] g (x)
g 0 (x)
g(x)
x
a
sinh x
cosh x
tanh x
cosech x
sech x
sech2 x
coth x
sinh2 x
cosh2 x
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f (x)dx
[g(x)]n+1
n+1
(n 6= −1)
ln |g (x)|
ax
(a > 0)
ln a
cosh x
sinh x
ln cosh x ln tanh x2 2 tan−1 ex
tanh x
ln |sinh x|
sinh 2x
− x2
4
sinh 2x
+ x2
4
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Section 4: Standard integrals
f (x)
1
a2 +x2
√ 1
a2 −x2
√
a2 − x2
9
R
f (x) dx
f (x)
1
a
tan−1
1
a2 −x2
R
(a > 0)
1
x2 −a2
f (x) dx
a+x 1
2a ln a−x (0 < |x| < a)
x−a 1
ln
2a
x+a (|x| > a > 0)
sin−1
x
a
√ 1
a2 +x2
√
2
2
ln x+ aa +x (a > 0)
(−a < x < a)
√ 1
x2 −a2
√
2
2
ln x+ xa −a (x > a > 0)
a2
2
−1
sin
√
+x
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x
a
x
a
a2 −x2
a2
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√
i √
a2 +x2
a2
2
x2 −a2
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a2
2
J
h
h
sinh−1
− cosh−1
I
x
a
i
√
x a2 +x2
2
a
i
√
2
2
+ x xa2−a
+
x
a
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Section 5: Tips
10
5. Tips
● STANDARD INTEGRALS are provided. Do not forget to use these
tables when you need to
● When looking at the THEORY, STANDARD INTEGRALS, ANSWERS or TIPS pages, use the Back button (at the bottom of the
page) to return to the exercises
● Use the solutions intelligently. For example, they can help you get
started on an exercise, or they can allow you to check whether your
intermediate results are correct
● Try to make less use of the full solutions as you work your way
through the Tutorial
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Solutions to exercises
11
Full worked solutions
Exercise 1.
R
cos 3x cos 2x dx:
Use cos A cos B =
1
2
[cos (A + B) + cos (A − B)]
i.e. taking A = 3x and B = 2x:
R
R
cos 3x cos 2x dx = 12 [cos (3x + 2x) + cos (3x − 2x)]
R
= 12 [cos 5x + cos x] dx
Each term under the integration sign is a function of a linear function
of x, i.e.
R
R
f (ax+b) dx = a1 f (u)du, where u = ax+b, du = a dx, i.e. dx = du
a .
R
1
i.e. cos 3x cos 2x dx = 21 15 sin 5x+ 21 sin x+C = 10
sin 5x+ 12 sin x+C.
Return to Exercise 1
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Solutions to exercises
12
Exercise 2.
R
sin 5x cos 3x dx:
Use sin A cos B =
1
2
[sin (A + B) + sin (A − B)]
i.e. taking A = 5x and B = 3x:
Z
Z
1
[sin (5x + 3x) + sin (5x − 3x)]
sin 5x cos 3x dx =
2
Z
1
[sin 8x + sin 2x] dx
=
2
i.e.
Z
sin 5x cos 3x dx = −
= −
11
11
cos 8x −
cos 2x + C
28
22
1
1
cos 8x − cos 2x + C.
16
4
Return to Exercise 2
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Solutions to exercises
13
Exercise 3.
R
sin 6x sin 4x dx:
Use sin A sin B = − 12 [cos (A + B) − cos (A − B)]
i.e. taking A = 6x and B = 4x:
Z
Z
1
[cos (6x + 4x) − cos (6x − 4x)]
sin 6x sin 4x dx = −
2
Z
1
[cos 10x − cos 2x] dx
= −
2
i.e.
Z
11
1 1
sin 6x sin 4x dx = −
sin 10x +
sin 2x + C
2 10
22
= −
1
1
sin 10x + sin 2x + C.
20
4
Return to Exercise 3
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Solutions to exercises
14
Exercise 4.
R
cos 2ωt sin ωt dt:
Use sin A cos B =
1
2
[sin (A + B) + sin (A − B)]
i.e. taking A = ωt and B = 2ωt:
Z
Z
1
[sin (ωt + 2ωt) + sin (ωt − 2ωt)]
cos 2ωt sin ωt dt =
2
Z
1
[sin 3ωt + sin (−ωt)] dt
=
2
Z
1
=
[sin 3ωt − sin ωt] dt
2
11
1 1
cos 3ωt +
cos ωt + C
= −
2 3ω
2ω
1
1
= −
cos 3ωt +
cos ωt + C.
6ω
2ω
Return to Exercise 4
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Solutions to exercises
15
Exercise 5.
R
cos 4ωt cos 2ωt dt:
Use cos A cos B =
1
2
[cos (A + B) + cos (A − B)]
i.e. taking A = 4ωt and B = 2ωt:
1
2
Z
cos 4ωt cos 2ωt dt =
1
2
Z
=
=
1 1
1 1
sin 6ωt +
sin 2ωt + C
2 6ω
2 2ω
=
1
1
sin 6ωt +
sin 2ωt + C.
12ω
4ω
Z
[cos (4ωt + 2ωt) + cos (4ωt − 2ωt)]
[cos 6ωt + cos 2ωt] dt
Return to Exercise 5
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Solutions to exercises
Exercise 6.
R
sin2 x dx:
For the particular case:
A=B=x,
sin A sin B = − 12 [cos (A + B) − cos (A − B)],
the formula:
reduces to:
16
sin2 x = − 12 (cos 2x − 1)
(a half-angle formula)
i.e.
Z
sin2 x dx = −
1
2
Z
(cos 2x − 1) dx
1
11
sin 2x + x + C
22
2
1
1
= − sin 2x + x + C.
4
2
= −
Return to Exercise 6
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Solutions to exercises
Exercise 7.
R
sin2 ωt dt:
For the particular case:
A=B=ωt,
sin A sin B = − 12 [cos (A + B) − cos (A − B)],
the formula:
reduces to:
17
sin2 ωt = − 12 (cos 2ωt − 1)
i.e.
Z
sin2 ωt dt = −
1
2
Z
(cos 2ωt − 1) dt
1
1 1
sin 2ωt + t + C
2 2ω
2
1
1
sin 2ωt + t + C.
= −
4ω
2
= −
Return to Exercise 7
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Solutions to exercises
Exercise 8.
R
cos2 t dt:
18
For the particular case:
the formula: cos A cos B =
reduces to:
1
2
A=B=t,
[cos (A + B) + cos (A − B)]
,
cos2 t = 21 (cos 2t + 1)
i.e.
Z
2
cos t dt
=
=
=
1
2
Z
(cos 2t + 1) dt
11
1
sin 2t + t + C
22
2
1
1
sin 2t + t + C.
4
2
Return to Exercise 8
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Solutions to exercises
Exercise 9.
R
cos2 kx dx:
19
For the particular case:
the formula: cos A cos B =
reduces to:
1
2
A=B=kx,
[cos (A + B) + cos (A − B)]
,
cos2 kx = 21 (cos 2kx + 1)
i.e.
Z
2
cos kx dx =
=
=
1
2
Z
(cos 2kx + 1)dx
1 1
1
sin 2kx + x + C
2 2k
2
1
1
sin 2kx + x + C.
4k
2
Return to Exercise 9
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