7
Techniques of Integration
,
.
Copyright © Cengage Learning. All rights reserved.
k
(gcr ) g태 t
S 5 Sf
l
=
~
xFglt )
f
g
7.3
74
=
:
-
1
남
~
따
A
Trigonometric Substitution
coshojl sinhxit
(
5im
2
비
+
cos ②
1 *
tau
1
+
cot θ
수
θ
sec @
=
=
csc θ
VFxz
FEn=
블
ㅡ
ㅡ
Fx
=
cosll
SiMO
Copyright © Cengage Learning. All rights reserved.
Trigonometric Substitution (1 of 8)
In finding the area of a circle or an ellipse, an integral of the form
arises, where a > 0.
Mmai
.
"
If the integral were
would be effective
but, as it stands,
e=
asinθ
Vaz
-
xz
:
=
=
F
,
- π
]
π
→
5
-
a
,
a
]
:
onelooue
az
의
asiuzorakos
-
acos θ
(
:
Cos
θ
xo onf *
파
,
z
J
‹#›
Trigonometric Substitution (2 of 8)
If we change the variable from x to ㅂ by the substitution x = a sin ㅂ , then the
identity
allows us to get rid of the root sign because
=
Q CosQ
0
‹#›
Trigonometric Substitution (3 of 8)
Notice the difference between the substitution
(in which the new variable
is a function of the old one) and the substitution x = a sinθ (the old variable is a
function of the new one).
In general, we can make a substitution of the form x = g(t) by using the
Substitution Rule in reverse.
To make our calculations simpler, we assume that g has an inverse function;
that is, g is one-to-one.
ㄷ
U
.
-
cF
Sigx
지
K르 t
kdn
=
S igciat
.
이
x
1
‹#›
Trigonometric Substitution (4 of 8)
In this case, if we replace u by x and x by t in the Substitution Rule, we obtain
This kind of substitution is called inverse substitution.
We can make the inverse substitution x = a sin θ provided that it defines a oneto-one function.
This can be accomplished by restricting θ to lie in the interval
‹#›
Trigonometric Substitution (5 of 8)
nC= tceh θ
"
Table of Trigonometric Substitutions
l
1
득
In each case the restriction on θ is imposed to ensure that the
function that defines the substitution is one-to-one.
‹#›
vxF
a>
o
K
- o ≤
aseeo
(
π
스
K
3
seco
osot
~
.
K
=
=
Cos θ
*
-
따
*
*
1
*
r
π
rrat
x
γ
px -ar ) asec @ =
=
ㅡ
ㅡ
=
이
a
t 때
atan
θ
어
K
T
on
<π
비스 <
3
피
π
Example 1
Evaluate
Solution:
Let x = 3 sin 비, where
Then dx = 3 cos θ dθ and
(Note that cos 비 ≥ 0 because
‹#›
Example 1 – Solution (1 of 4)
Thus the Inverse Substitution Rule gives
x
3 sin ② (
=
x
3
-
=
따쁘드파
STu
비
5
θ=
1
+
srule
coA급
)
=
)
입
Csc
coto
cse θ
)
'
=
-
cof ( siul
5 )
x
=
-
-
siv
"
(
k
5 ) tc
‹#›
Example 1 – Solution (2 of 4)
Because this is an indefinite integral, we must return to the original variable x.
This can be done either by using trigonometric identities to express cot θ in terms of
,
or by drawing a diagram, as in Figure 1, where θ is interpreted as an
맺
angle of a right triangle.
C
=
*
:
3 SM @
L
E θ 스π
O
≤ S
cot θ=
파
xar
a
-
x
"ii
Other
eases
?
Figure 1
‹#›
3
x
snaciaim
=
luk
-
~
x2
-
M )
≤ 비드
)
Cot =
⇒
a
( *I
x
↓≤≤ KE
한
0
cof
:
-"조C
π
=
⇒
-x
is
<
3
o
o
'
Ko θ
≤
우 xKo aad
=
cost
파
35
비=
sin t -
nt = 3
=
=
x
)
-
cost
-
θ
=
sint
v
"
sin <
q
-
C
'
l)
'
그
-
대
3 siu )
cofa
-
-
느ㅠ파
L dome !
true !
=벼
cot θ
xa
=
ㅠ
ton
k
≤θ
C= 35
eot
+=
=
11
c*
일
t
⇒
"
par
-
π
θ
'
)
~
xa
a
- xp
-
=
-
KC )
3 e
-
7
heletore
,
for
cstrue
Al
θ
E
(
i-
[
ㅠ
교]
"
)
Example 1 – Solution (3 of 4)
Since
we label the opposite side and the hypotenuse as having
C
빗번
lengths x and 3.
Then the Pythagorean Theorem gives the length of the adjacent side as
)
so we can simply read the value of cot θ from the figure:
3
x
바
9
~2
9 -C
(Although θ > 0 in the diagram, this expression for cot
0.)
t
-
다
.
c
0
=
θ= x
"
os
0
.
Fsin
=
.
젤
15
=
5
.
F
;
Q is
valid even when θ K
.
‹#›
Example 1 – Solution (4 of 4)
Since
we have
‹#›
Example 2
Find the area enclosed by the ellipse 타원
b
-
Solution:
Solving the equation of the ellipse for y, we get
≈
ara
르짜
5
사
a
>
-
4-
5
까
arn
k
‹#›
Example 2 – Solution (1 of 5)
Because the ellipse is symmetric with respect to both axes, the total area A is
four times the area in the first quadrant (see Figure 2).
Figure 2
‹#›
Example 2 – Solution (2 of 5)
The part of the ellipse in the first quadrant is given by the function
and so
To evaluate this integral we substitute x = a sin θ. Then dx = a cos θ dθ.
‹#›
Example 2 – Solution (3 of 5)
To change the limits of integration we note that when x = 0, sin θ = 0, so θ = 0;
when x = a, sin θ = 1, so
Also
since
‹#›
Example 2 – Solution (4 of 5)
Therefore
‹#›
Example 2 – Solution (5 of 5)
We have shown that the area of an ellipse with semiaxes a and b is πab.
In particular, taking a = b = r, we have proved the famous formula that the area
of a circle with radius r is
Note:
Because the integral in Example 2 was a definite integral, we
changed the limits of integration and did not have to convert back
to the original variable x.
‹#›
Example 3
1
Find
sut
+
x
은
←
r
rπ
4
Solution:
Let x = 2
cosht
=
sTyht
rFsinh
'
=
e
zco .
+4
oshtzoshttz
=
lAS
.
immzt
째도
"
*
.
4
.
5
냐따만
… )
‹#›
Example 3 – Solution (1 of 3)
So we have
To evaluate this trigonometric integral we put everything in terms of sin θ and
cos θ :
‹#›
Example 3 – Solution (2 of 3)
Therefore, making the substitution u = sin θ, we have
- claim:Sn
sino
cosec θ=
=
siwoseco
Sec
=
쳤에
CoE②
θ
=
9C
=
atan θ
때비
에다
xitr
=
"
sK
ㅮ
9
오
mθ
5
itnro
=
용
2t
4
ㅠ
<
Qo
Cohen
a
)
=
+4
~
xx
‹#›
Example 3 – Solution (3 of 3)
We use Figure 3 to determine that
and so
Figure 3
‹#›
Example 5
Evaluate
9
Solution 1:
We let x = a sec θ, where
s
←
졌고
양깨따
;
Then dx = a sec θ tan θ dθ and
%
‹#›
Example 5 – Solution 1 (2 of 3)
so we have
e=
asec @
⑤=
답일
c0 s
크피로
:
캠
:
The triangle in Figure 4 gives
파
oCO
<π
등
KEa
θ
ean
=
Figure 4
‹#›
( a 70 )
Claine
i
=
c
π
asecQ
L θ< EE
xz
apr
π KT
-
tan θ
⇒
=
π .θ
C
=
아
223
)
π
a
=π
⇒
tan
=
쁘
K
≥
x
ar
3 Q
a
T π
고간
비=
Oc =
3
L
C
π
θㅠ
:
자비
ㅠ
ur
e
asecc
sec ② or
=
=
음
s
수e
1- >
asec
=
θ
,
Then
tan
Koc
>
ㅡ
-
'
<π
V
t
=
a
tn θ
(
π+ 비
놓
"
'
θ
t예저비
=
' E π
oL
and θ
때어까
=
-
)
음
s θ
<
,
)
+rraby
암
파
-
ㅁ
o
Example 5 – Solution 1 (3 of 3)
Writing C1 = C − In a, we have
‹#›
Example 5 – Solution 2 (1 of 2)
For x > 0 the hyperbolic substitution x = a cosh t can also be used.
Using the identity
se
=
we have
acosht
~
coshzt
=
a
sKuhp 디
-
) siult /
.
>
아
t
[
o
,
∞
)
F,
→
18
수
∞
)
1
o.
∞
y
suht
=
st
)
-
→
[
0
.
∞
)
‹#›
Example 5 – Solution 2 (2 of 2)
Since dx = a sinh t dt, we obtain
oe←
cosut
a
coshst
siuh =
로
-
수
Since
‹#›
Trigonometric Substitution (8 of 8)
Note:
As Example 5 illustrates, hyperbolic substitutions can be used in place of
trigonometric substitutions and sometimes they lead to simpler answers.
But we usually use trigonometric substitutions because trigonometric identities
are more familiar than hyperbolic identities.
‹#›
Example 6
Find
91
4
%+
Solution:
?
를
3
J
차
4 pL
IC
+
=
x
량
=
axx R
⇒
4
.
x
13
고
2
(Cor
:
=
-
cR p ]
23
3
k
an
」
Although
is not quite one of the expressions in the table of trigonometric
substitutions, it becomes one of them if we make the preliminary substitution
u = 2x, which gives
ㅡ
‹#›
Example 6 – Solution (1 of 3)
Then we substitute
which
gives
(
-
Z
π
ㅠ
cQc
)
.
많좋
"
When x = 0, tan Q = 0, so O = 0;
when
‹#›
Example 6 – Solution (2 of 3)
높
=
rupe vS .dui
-
r
Now we substitute u = cos θ so that du = –sin θ dθ. When θ = 0, u = 1; when
롭지.
.
고
.
‹#›
Example 6 – Solution (3 of 3)
Therefore
‹#›
V
4 주
x
β
9
C T(
~→
K=
3
-
2
tc@
32
v
x + &x
>
+
주KI.
)( 2
K 1
V
~
4x4 3
4
=
x
고
teu) D
+
5
lx + 3 8 pL
-
3
)
-
A
=
2