Download 1/24/2011

1/24/2011
6.2 Trigonometric Integrals and Trigonometric Substitution
Memorize the following trig identities:
cos2x + sin
+ i 2x = 1
1
cos2x = 1 ‐ sin2x sin2x = 1 ‐ cos2x sec2x = 1 + tan2x
tan2x = sec2x ‐ 1
cos 2 x =
1
[1 + cos 2 x]
2
1
[1 − cos 2 x]
2
sin 2x = 2sinxcosx
sin 2 x =
1 + cot2x = csc2x
Review of Trigonometric Derivatives
D(sin x) = cosx
∫ cos xdx = sin x + C
D(cos x) = ‐sinx
∫ sin xdx = − cos x + C
D(tanx) = sec2x
∫ sec
D(cot x) = ‐csc2x
∫ csc
D(sec x) = secx tanx
∫ sec x tan xdx = sec x + C
D(csc x) = ‐cscx cotx
∫ csc x cot xdx = − csc x + C
2
xdx = tan x + C
2
xdx = − cot x + C
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Case I: Integrating with odd powers of sine or cosine.
∫ sin x cos
5
Example 2: Evaluate
2
xdx
Strategy: split off a single factor of cosx or sinx (whichever is odd)
=∫ sin4 x ⋅ sin x ⋅ cos2 xdx
Use the identity sin
y 2x = 1 – cos2x
Note: sin4x = (sin2x)2 = (1‐cos2x)2
=∫ (1 − cos2 x )2 ⋅ sin x ⋅ cos2 xdx
= ∫ (1 − cos2 x )2 cos2 x ⋅ sin xdx
Then use a u-substitution:
Let u = cos x → du = -sinxdx → -du = sinxdx
→ − ∫ (1 - 2u2 + u4 ) ⋅ u2du
Substituting: - ∫ (1
(1-u
u2 )2 ⋅ u2du
4
6
u3 2u 5 u 7
=- +
−
+C
3
5
7
=
C
+
x
7
s
o 7
c
x
5
s
o 5
c
2
+
x
3
s
o 3
c
-
= - ∫ (u − 2u + u )du
2
Case II: Sine/Cosine have even powers
1
2
1
2
Use a half‐angle formula cos2 x = (1 + cos(2 x )) ; sin2 x = (1 − cos(2 x ))
π
π
= ∫ [sin2 (3t )]2 dt
Evaluate : ∫ sin (3t )dt
4
0
0
π
1
= ∫ [ (1 − cos(6t ))]2 dt
2
0
π
=
1
∫ [ 4 (1 − 2cos(6t ) + cos (6t )]dt
2
0
π
π
∫
t
d
t
6
π
1
1
1dt − ∫ 2cos(6t )dt +
∫
40
40
2
s
o
c
0
π
1 4
1
= ∫ (1 − 2cos(6t ) + cos2 (6t ))dt
40
︵ ︶
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+
0
∫ [1 + cos(12t )]dt =
0
0
π
sin(12t )
[t +
]
12
π
0
1
sin(12π )
sin0
[π +
−0−
]=
8
12
12
=
π
3 8
⋅0 +
=
8
2
1 1
1 4
⋅π −
=
π
︶]dt
π 8
1 8
sin(12t )
[t +
]
12
︵
0
1 8
1 8
π
∫
t
2
1
π
s
o
c
+
1
1[2
0
π
1 4
1
1 ⎡ 2sin(6t ) ⎤
= [t ] −
⎥⎦
4
4⎢
6
⎣
π
Integrating even powers of tangent and secant:
Strategy: Factor out sec2x and use the identity 1 + tan2x = sec2x.
∫ tan
6
x sec 4 xdx = ∫ tan 6 x ⋅ sec 2 x ⋅ sec 2 xdx
= ∫ tan 6 x(1 + tan 2 x) sec 2 xdx
6
((1 + u 2 )du = ∫ (u 6 + u 8 )du
Ans :
C
+
x
9
9
n
a
t
+
x
7
7
n
a
t
∫u
3
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Integrating with odd powers of tangent:
Strategy: Save a factor of secxtanx and use tan2x = sec2x – 1 to
express the remaining factors of secx. Then use u‐substitution.
t (2 x) sec (2 x) ⋅ sec(2
(2 x) tan(2
t (2 x))ddx
t (2 x) sec (2 x))ddx = ∫ tan
∫ tan
3
5
2
4
= ∫ (sec 2 (2 x) − 1) sec 4 (2 x) ⋅ sec(2 x) tan(2 x)dx
Let u = sec(2x)
du = 2sec(2x)tan(2x)dx
½ du = sec(2x)tan(2x)dx
½ du = sec(2x)tan(2x)dx
C
+
x
2
0
5 1
c
e
s
x
2
4
7 1
c
e
s
:
s
n
A
1
2
4 1
(
u
−
1)
u
⋅
du
=
(u 6 − u 4 )du
∫
∫
2
2
︵
︶
4
∫
︵
c
e
s
π /2
︶
(t / 2)dt
0
8 3
:
s
n
A
4
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2
x
Using Trigonometric Substitutions
∫
dx
Is a the hypotenuse or a leg?
9 − x2
a=3
x = 3sinθ
dx = 3cosθdθ
x2 = 9sin2 θ
x
θ
9 − x2
∫
∫
9sin 2 θ
9 − 9sin 2 θ
⋅ 3cos θ dθ = ∫
27 sin 2 θ
9(1 − sin θ )
2
⋅ cos θ dθ
27 sin 2 θ
2
⋅ cos θ dθ = ∫ 27 sin θ ⋅ cos θ dθ
3cos θ
9 cos 2 θ
1
2
9sin
9
i
θ
d
θ
=
9
i (2θ )]dθ
∫
∫ 2[(1 − sin(2
9
1
[θ + cos(2θ )] + C
2
2
a=3
θ
9 − x2
2
⎛ x⎞
⎝ ⎠
x
- 3
9
Now back substitute to get
the answer:
θ = sin −1 ⎜ ⎟
3
x
What is cosθ ?
cos (2θ ) = 2sinθcosθ
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9
x 9 − x2
−1 ⎛ x ⎞
= [sin ⎜ ⎟ −
]+C
2
3
⎝3⎠ 3
C
+
2
x
9
x 2
x 3
n
i
s
9 2
:
s
n
A
⎛ ⎞
⎜ ⎟
⎝ ⎠
1
-
=
x
∫
x −7
2
dx
Strategy?
u-substitution
Ans : x 2 − 7 + C
∫
x −9
dx
x3
Ans :
1
⎛x⎞
sec −1 ⎜ ⎟ −
6
⎝3⎠
use x = 3secθ
Strategy?
2
x2 − 9
x
x2 − 9
+C
2x 2
θ
3
6