Trigonometric Substitution
In finding the area of a circle or an ellipse, an integral of the form x sa 2 x 2 dx arises,
where a 0. If it were x xsa 2 x 2 dx, the substitution u a 2 x 2 would be effective
but, as it stands, x sa 2 x 2 dx is more difficult. If we change the variable from x to by
the substitution x a sin , then the identity 1 sin 2 cos 2 allows us to get rid of the
root sign because
sa 2 x 2 sa 2 a 2 sin 2 sa 21 sin 2 sa 2 cos 2 a cos Notice the difference between the substitution u a 2 x 2 (in which the new variable is
a function of the old one) and the substitution x a sin (the old variable is a function of
the new one).
In general we can make a substitution of the form x tt by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that t has an inverse function; that is, t is one-to-one. In this case, if we replace u by x and x by t in the Substitution
Rule (Equation 5.5.4), we obtain
y f x dx y f tttt dt
This kind of substitution is called inverse substitution.
We can make the inverse substitution x a sin provided that it defines a one-to-one
function. This can be accomplished by restricting to lie in the interval 2, 2.
In the following table we list trigonometric substitutions that are effective for the given
radical expressions because of the specified trigonometric identities. In each case the restriction on is imposed to ensure that the function that defines the substitution is one-to-one.
(These are the same intervals used in Appendix D in defining the inverse functions.)
Table of Trigonometric Substitutions
Expression
Substitution
Identity
sa 2 x 2
x a sin ,
2
2
1 sin 2 cos 2
sa 2 x 2
x a tan ,
2
2
1 tan 2 sec 2
sx 2 a 2
x a sec ,
0
EXAMPLE 1 Evaluate
y
3
or 2
2
sec 2 1 tan 2
s9 x 2
dx.
x2
SOLUTION Let x 3 sin , where 2 2. Then dx 3 cos d and
s9 x 2 s9 9 sin 2 s9 cos 2 3 cos 3 cos (Note that cos 0 because 2 2.) Thus, the Inverse Substitution Rule
gives
3 cos s9 x 2
y x 2 dx y 9 sin 2 3 cos d
y
cos 2
d y cot 2 d
sin 2
y csc 2 1 d
cot C
1
2 ■ TRIGONOMETRIC SUBSTITUTION
Since 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 in terms of sin x3 or
by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
Since sin x3, we label the opposite side and the hypotenuse as having lengths x and 3.
Then the Pythagorean Theorem gives the length of the adjacent side as s9 x 2, so we
can simply read the value of cot from the figure:
3
x
¨
œ„„„„„
9-≈
FIGURE 1
sin ¨=
cot x
3
s9 x 2
x
(Although 0 in the diagram, this expression for cot is valid even when 0.)
Since sin x3, we have sin1x3 and so
y
x
s9 x 2
s9 x 2
dx sin1
2
x
x
3
C
EXAMPLE 2 Find the area enclosed by the ellipse
x2
y2
1
a2
b2
y
SOLUTION Solving the equation of the ellipse for y, we get
(0, b)
y2
x2
a2 x2
1
b2
a2
a2
(a, 0)
0
x
y
or
b
sa 2 x 2
a
FIGURE 2
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). The part of the ellipse in the first quadrant is
given by the function
b
y sa 2 x 2
0xa
a
¥
≈
+ =1
b@
a@
and so
1
4
Ay
a
0
b
sa 2 x 2 dx
a
To evaluate this integral we substitute x a sin . Then dx a cos d. To change the
limits of integration we note that when x 0, sin 0, so 0; when x a,
sin 1, so 2. Also
sa 2 x 2 sa 2 a 2 sin 2 sa 2 cos 2 a cos a cos since 0 2. Therefore
A4
b
a
y
a
0
4ab y
2
0
[
sa 2 x 2 dx 4
b
a
cos 2 d 4ab y
2ab 12 sin 2
y
2 1
2
0
2
0
]
2
0
2ab
a cos a cos d
1 cos 2 d
00
2
ab
We have shown that the area of an ellipse with semiaxes a and b is ab. In particular,
taking a b r, we have proved the famous formula that the area of a circle with
radius r is r 2.
NOTE Since 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.
■
TRIGONOMETRIC SUBSTITUTION ■ 3
EXAMPLE 3 Find
1
y x sx
2
2
4
dx.
SOLUTION Let x 2 tan , 2 2. Then dx 2 sec 2 d and
sx 2 4 s4tan 2 1 s4 sec 2 2 sec 2 sec Thus, we have
dx
y x sx
2
y
4
2
2 sec 2 d
1
2
4 tan 2 sec 4
sec y tan d
2
To evaluate this trigonometric integral we put everything in terms of sin and cos :
sec
1
cos 2
cos 2
2
tan cos sin sin 2
Therefore, making the substitution u sin , we have
y
dx
1
2
x sx 4
4
2
œ„„„„„
≈+4
1
4
x
¨
2
FIGUR E 3
tan ¨=
cos 1
d 2
sin 4
y
1
u
y
C
du
u2
1
C
4 sin csc C
4
We use Figure 3 to determine that csc sx 2 4x and so
x
2
dx
y x sx
2
EXAMPLE 4 Find
y sx
2
4
sx 2 4
C
4x
x
dx.
4
2
SOLUTION It would be possible to use the trigonometric substitution x 2 tan here (as in
Example 3). But the direct substitution u x 2 4 is simpler, because then du 2x dx
and
y
NOTE
x
1
dx 2
sx 4
2
du
y su
su C sx 2 4 C
Example 4 illustrates the fact that even when trigonometric substitutions are possible, they may not give the easiest solution. You should look for a simpler method first.
■
EXAMPLE 5 Evaluate
y sx
dx
, where a 0.
a2
2
SOLUTION We let x a sec , where 0 dx a sec tan d and
2 or 32. Then
sx 2 a 2 sa 2sec 2 1 sa 2 tan 2 a tan a tan Therefore
y sx
dx
a sec tan y
d
2
a
a tan 2
y sec d ln sec tan C
4 ■ TRIGONOMETRIC SUBSTITUTION
The triangle in Figure 4 gives tan sx 2 a 2a, so we have
x
≈-a@
œ„„„„„
y sx
¨
sec ¨=
x
a
ln x sx 2 a 2 ln a C
a
FIGU RE 4
dx
x
sx 2 a 2
ln
C
2
a
a
a
2
Writing C1 C ln a, we have
y sx
EXAMPLE 6 Find
y
3 s32
0
dx
ln x sx 2 a 2 C1
a2
2
x3
dx.
4x 2 932
SOLUTION First we note that 4x 2 932 s4x 2 9
)3 so trigonometric substitution
is appropriate. Although s4x 2 9 is not quite one of the expressions in the table of
trigonometric substitutions, it becomes one of them if we make the preliminary substitu3
tion u 2x. When we combine this with the tangent substitution, we have x 2 tan ,
3
2
which gives dx 2 sec d and
s4x 2 9 s9 tan 2 9 3 sec When x 0, tan 0, so 0; when x 3s32, tan s3, so 3.
y
3 s32
0
27
3
x3
3 8 tan dx
y
2
32
0
4x 9
27 sec3
163 y
sec 2 d
3
tan 3
3 sin d 163 y
d
0
sec cos2
3
0
163 y
3
2
1 cos 2
sin d
cos 2
3
0
Now we substitute u cos so that du sin d. When 0, u 1; when
3, u 12.
Therefore
y
3 s32
0
2
12 1 u
12
x3
3
3
dx
du 16 y 1 u 2 du
y
16
2
32
2
1
1
4x 9
u
163 u EXAMPLE 7 Evaluate
x
y s3 2x x
2
1
u
12
163 [( 12 2) 1 1] 323
1
dx.
SOLUTION We can transform the integrand into a function for which trigonometric substi-
tution is appropriate by first completing the square under the root sign:
3 2x x 2 3 x 2 2x 3 1 x 2 2x 1
4 x 12
This suggests that we make the substitution u x 1. Then du dx and x u 1, so
x
y s3 2x x
2
dx y
u1
du
s4 u 2
TRIGONOMETRIC SUBSTITUTION ■ 5
We now substitute u 2 sin , giving du 2 cos d and s4 u 2 2 cos , so
x
y s3 2x x
2
2 sin 1
2 cos d
2 cos dx y
y 2 sin 1 d
2 cos C
s4 u 2 sin1
u
2
C
s3 2x x 2 sin1
x1
2
C
Exercises
A Click here for answers.
Click here for solutions.
S
y
21.
y
23.
y s5 4x x
25.
y s9x
27.
y x
29.
y x s1 x
1–3
Evaluate the integral using the indicated trigonometric
substitution. Sketch and label the associated right triangle.
1.
2.
3.
yx
yx
1
dx ;
sx 2 9
x 3 sec s9 x 2 dx ;
x 3 sin 2
3
y sx
■
x3
dx ;
9
2
■
■
4–30
x 3 tan ■
■
■
■
■
■
■
■
Evaluate the integral.
4.
y
5.
y
2 s3
0
2
s2
3
x
dx
s16 x 2
1
dt
t 3 st 2 1
1
dx
x 2 s25 x 2
7.
y
9.
y sx
dx
16
11.
y s1 4x
13.
6.
8.
y
2
0
x 3 sx 2 4 dx
y
y st
2
2
1
t5
dt
2
dx
1
dx
6x 8
dx
2x 22
■
■
y
sx 2 9
dx
x3
14.
y u s5 u
15.
y
x2
dx
2
a x 2 32
16.
y
17.
y sx
x
dx
7
18.
y ax
4
dx
■
■
y sx
22.
y
24.
y st
26.
y s4x x
28.
y 5 4x x
30.
y
1
0
2
dt
sx 2 1 dx
2
dt
6t 13
x2
2
dx
dx
■
■
cos t
dt
s1 sin 2 t
2
0
■
■
■
dx
b 2 32
■
dx
ln ( x sx 2 a 2 ) C
a2
y sx
dx
x
sinh1
a2
a
2
C
2
dx
x 2 s16x 2 9
■
2
These formulas are connected by Formula 3.9.3.
du
2 52
(b) Use the hyperbolic substitution x a sinh t to show that
2
2
y s25 t
31. (a) Use trigonometric substitution to show that
x sx 4 dx
y
0
2
2
12.
2
x 3s4 9x 2 dx
sx 2 a 2
dx
x4
dx
2
23
0
t
20.
■
■
10.
2
■
s1 x 2
dx
x
19.
32. Evaluate
y x
2
x2
dx
a 2 32
(a) by trigonometric substitution.
(b) by the hyperbolic substitution x a sinh t.
■
6 ■ TRIGONOMETRIC SUBSTITUTION
33. Find the average value of f x sx 2 1x, 1 x 7.
34. Find the area of the region bounded by the hyperbola
9x 2 4y 2 36 and the line x 3.
where is the charge density per unit length on the rod and 0
is the free space permittivity (see the figure). Evaluate the integral to determine an expression for the electric field EP.
y
35. Prove the formula A 2 r 2 for the area of a sector of a circle
1
with radius r and central angle . [Hint: Assume 0 2
and place the center of the circle at the origin so it has the
equation x 2 y 2 r 2. Then A is the sum of the area of the
triangle POQ and the area of the region PQR in the figure.]
y
P (a, b)
0
L
x
P
39. Find the area of the crescent-shaped region (called a lune)
bounded by arcs of circles with radii r and R. (See the figure.)
¨
O
Q
R
x
; 36. Evaluate the integral
r
R
y
dx
x 4 sx 2 2
Graph the integrand and its indefinite integral on the same
screen and check that your answer is reasonable.
; 37. Use a graph to approximate the roots of the equation
x 2 s4 x 2 2 x. Then approximate the area bounded by
the curve y x 2 s4 x 2 and the line y 2 x.
38. A charged rod of length L produces an electric field at point
Pa, b given by
EP y
La
a
b
dx
4 0 x 2 b 2 32
40. A water storage tank has the shape of a cylinder with diameter
10 ft. It is mounted so that the circular cross-sections are vertical. If the depth of the water is 7 ft, what percentage of the
total capacity is being used?
41. A torus is generated by rotating the circle x 2 y R2 r 2
about the x-axis. Find the volume enclosed by the torus.
TRIGONOMETRIC SUBSTITUTION ■ 7
Answers
S
19. ln (s1 x 2 1)x s1 x 2 C
Click here for solutions.
21.
23.
sin1x 23 x 2s5 4x x 2 C
1
3
ln 3x 1 s9x 2 6x 8 C
1
2
1. sx 2 99x C
3. x 2 18 sx 2 9 C
25.
5. 24 s38 7. s25 x 225x C
27. 2 tan1x 1 x 1x 2 2x 2 C
1
4
1
3
9. ln (sx 2 16 x) C
11. 4 sin12x 2 x s1 4x 2 C
1
1
2
13. 6 sec x3 sx 92x 2 C
15. (xsa 2 x 2 ) sin1xa C
17. sx 2 7 C
1
1
64
1215
9
2
1
29.
1
4
sin1 x 2 14 x 2 s1 x 4 C
33.
1
6
(s48 sec1 7)
37. 0.81, 2; 2.10
39. r sR r r 2 R 2 arcsinrR
2
2
2
41. 2 2Rr 2
8 ■ TRIGONOMETRIC SUBSTITUTION
Solutions: Trigonometric Substitution
1. Let x = 3 sec θ, where 0 ≤ θ <
π
2
or π ≤ θ <
3π
2 .
Then
dx = 3 sec θ tan θ dθ and
p
p
√
p
x2 − 9 = 9 sec2 θ − 9 = 9(sec2 θ − 1) = 9 tan2 θ
= 3 |tan θ| = 3 tan θ for the relevant values of θ.
Z
1
√
dx =
x2 x2 − 9
Z
1
3 sec θ tan θ dθ =
9 sec2 θ · 3 tan θ
1
9
R
cos θ dθ =
1
9
Note that − sec(θ + π) = sec θ, so the figure is sufficient for the case π ≤ θ <
3. Let x = 3 tan θ, where − π2 < θ <
p
Z
x2 + 9 =
p
9 tan2 θ + 9 =
π
2.
q
sin θ + C =
1
9
√
x2 − 9
+C
x
3π
.
2
Then dx = 3 sec2 θ dθ and
9(tan2 θ + 1) =
√
9 sec2 θ
= 3 |sec θ| = 3 sec θ for the relevant values of θ.
Z
Z
33 tan3 θ
3 sec2 θ dθ = 33 tan3 θ sec θdθ = 33 tan2 θ tan θ sec θ dθ
3 sec θ
R¡ 2
¢
¢
3R ¡
=3
sec2 θ − 1 tan θ sec θ dθ = 33
u − 1 du
[u = sec θ, du = sec θ tan θ dθ]
#
" ¡
¢3/2
√
2
¢
¢
x2 + 9
3¡ 1 3
3¡ 1
3
3 1 x +9
−
+C
= 3 3 u − u + C = 3 3 sec θ − sec θ + C = 3
3
33
3
x3
√
dx =
x2 + 9
Z
=
1
3
¡
x2 + 9
¢3/2
−9
p
x2 + 9 + C
5. Let t = sec θ, so dt = sec θ tan θ dθ, t =
Z
2
√
2
Z
√
2
1
3
or
⇒ θ=
π
,
4
¡
¢p
x2 − 18
x2 + 9 + C
and t = 2 ⇒ θ =
π
.
3
Then
Z π/3
Z π/3
1
1
cos2 θ dθ
sec θ tan θ dθ =
dθ =
2
θ
tan
θ
sec
θ
π/4
π/4
π/4
R π/3 1
£
¤π/3
1
1
= π/4 2 (1 + cos 2θ) dθ = 2 θ + 2 sin 2θ π/4
h³
³
´
√
√
√ ´
¢i
¡
π
π
= 12 π3 + 12 23 − π4 + 12 · 1 = 12 12
+ 43 − 12 = 24
+ 83 − 14
1
√
dt =
t3 t2 − 1
π/3
sec3
7. Let x = 5 sin θ, so dx = 5 cos θ dθ. Then
Z
Z
1
1
√
5 cos θ dθ
dx =
52 sin2 θ · 5 cos θ
x2 25 − x2
R
1
1
csc2 θ dθ = − 25
cot θ + C
= 25
√
1 25 − x2
=−
+C
25
x
TRIGONOMETRIC SUBSTITUTION ■ 9
9. Let x = 4 tan θ, where − π2 < θ <
π
.
2
Then dx = 4 sec2 θ dθ and
p
√
√
x2 + 16 = 16 tan2 θ + 16 = 16(tan2 θ + 1)
√
= 16 sec2 θ = 4 |sec θ|
= 4 sec θ for the relevant values of θ.
Z
Z
4 sec2 θ dθ
= sec θ dθ = ln |sec θ + tan θ| + C1
4 sec θ
¯
¯√
¯ x2 + 16
¯√
¯
x¯
+ ¯¯ + C1 = ln ¯ x2 + 16 + x¯ − ln |4| + C1
= ln ¯¯
4
4
¡√
¢
= ln x2 + 16 + x + C, where C = C1 − ln 4.
dx
√
=
x2 + 16
(Since
Z
√
x2 + 16 + x > 0, we don’t need the absolute value.)
11. Let 2x = sin θ, where − π2 ≤ θ ≤
π
.
2
√
dx = 12 cos θ dθ, and 1 − 4x2 =
q
Then x =
1
2
sin θ,
1 − (2x)2 = cos θ.
¡
¢
R√
R
R
1 − 4x2 dx = cos θ 12 cos θ dθ = 14 (1 + cos 2θ) dθ
¡
¢
= 14 θ + 12 sin 2θ + C = 14 (θ + sin θ cos θ) + C
h
i
p
= 14 sin−1 (2x) + 2x 1 − 4x2 + C
13. Let x = 3 sec θ, where 0 ≤ θ < π2 or π ≤ θ < 3π
. Then
2
√
dx = 3 sec θ tan θ dθ and x2 − 9 = 3 tan θ, so
Z
Z
Z √ 2
x −9
3 tan θ
1
tan2 θ
dx
=
3
sec
θ
tan
θ
dθ
=
dθ
3
3
x
27 sec θ
3
sec2 θ
R
sin2 θ dθ =
=
1
3
=
³x´ 1
1
sec−1
−
6
3
6
15. Let x = a sin θ, where − π2 ≤ θ ≤
Z
x2 dx
(a2 − x2 )3/2
=
=
Z
Z
1
3
π
2.
R
1
(1
2
− cos 2θ) dθ = 16 θ −
sin 2θ + C = 16 θ −
1
6
sin θ cos θ + C
√
³ x ´ √x2 − 9
x2 − 9 3
1
+C
+ C = sec−1
−
x
x
6
3
2x2
Then dx = a cos θ dθ and
a2 sin2 θ a cos θ dθ
=
a3 cos3 θ
¡
1
12
Z
tan2 θ dθ
¢
sec2 θ − 1 dθ = tan θ − θ + C
x
x
=√
− sin−1 + C
a
a2 − x2
17. Let u = x2 − 7, so du = 2x dx. Then
Z
1
x
√
dx =
2
x2 − 7
Z
1
√ du =
u
1
2
·2
p
√
u + C = x2 − 7 + C.
10 ■ TRIGONOMETRIC SUBSTITUTION
19. Let x = tan θ, where − π2 < θ < π2 . Then dx = sec2 θ dθ
√
and 1 + x2 = sec θ, so
Z
Z √
Z
1 + x2
sec θ
sec θ
dx =
sec2 θ dθ =
(1 + tan2 θ) dθ
x
tan θ
tan θ
R
= (csc θ + sec θ tan θ) dθ
= ln |csc θ − cot θ| + sec θ + C [by Exercise 39 in Additional Topics: Trigonometric Integrals]
¯ √
¯√
¯
¯√
¯ 1 + x2 − 1 ¯ √
¯ 1 + x2
1 + x2
1 ¯¯
¯
¯ + 1 + x2 + C
¯
− ¯+
+ C = ln ¯
= ln ¯
¯
x
x
1
x
21. Let u = 4 − 9x2
⇒
R 2/3
0
du = −18x dx. Then x2 =
1
9
(4 − u) and
√
¡ 1¢
R0
du =
x3 4 − 9x2 dx = 4 91 (4 − u)u1/2 − 18
=
1
162
Or: Let 3x = 2 sin θ, where − π2 ≤ θ ≤
h
8 3/2
3u
− 25 u5/2
i4
=
0
1
162
1
162
£ 64
3
R4³
−
0
64
5
´
4u1/2 − u3/2 du
¤
=
64
1215
π
.
2
23. 5 + 4x − x2 = −(x2 − 4x + 4) + 9 = −(x − 2)2 + 9. Let
x − 2 = 3 sin θ, − π2 ≤ θ ≤ π2 , so dx = 3 cos θ dθ. Then
R√
Rp
Rp
5 + 4x − x2 dx =
9 − (x − 2)2 dx =
9 − 9 sin2 θ 3 cos θ dθ
√
R
R
=
9 cos2 θ 3 cos θ dθ = 9 cos2 θ dθ
R
¡
¢
= 92 (1 + cos 2θ) dθ = 92 θ + 12 sin 2θ + C
= 92 θ +
=
9
2
=
9
2
9
4
sin 2θ + C = 92 θ + 94 (2 sin θ cos θ) + C
√
µ
¶
5 + 4x − x2
9 x−2
−1 x − 2
sin
+ ·
·
+C
3
2
3
3
¶
µ
√
1
x−2
sin−1
+ (x − 2) 5 + 4x − x2 + C
3
2
let u = 3 sec θ, where 0 ≤ θ <
Z
1
du
√3
=
u2 − 9
Z
=
1
3
π
2
Z
Z
1
du
dx
√
√3
=
. Now
2
9x + 6x − 8
u2 − 9
√
or π ≤ θ < 3π
. Then du = 3 sec θ tan θ dθ and u2 − 9 = 3 tan θ, so
2
25. 9x2 + 6x − 8 = (3x + 1)2 − 9, so let u = 3x + 1, du = 3dx. Then
√
¯
¯ u + u2 − 9
R
sec θ tan θ dθ
= 13 sec θdθ = 13 ln|sec θ + tan θ| + C1 = 13 ln¯¯
3 tan θ
3
¯
¯
¯
¯
p
p
¯
¯
¯
¯
ln¯u + u2 − 9¯ + C = 13 ln¯3x + 1 + 9x2 + 6x − 8 ¯ + C
¯
¯
¯ + C1
¯
27. x2 + 2x + 2 = (x + 1)2 + 1. Let u = x + 1, du = dx. Then
Z
dx
=
(x2 + 2x + 2)2
Z
R
du
=
(u2 + 1)2
Z
sec2 θdθ
sec4 θ
"
where u = tan θ, du = sec2 θ dθ,
and u2 + 1 = sec2 θ
#
R
cos2 θ dθ = 12 (1 + cos 2θ) dθ = 12 (θ + sin θ cos θ) + C
¸
·
·
¸
1
1
u
x+1
−1
−1
+C =
tan u +
tan (x + 1) + 2
+C
=
2
1 + u2
2
x + 2x + 2
=
TRIGONOMETRIC SUBSTITUTION ■ 11
29. Let u = x2 , du = 2x dx. Then
R
√
¡
R√
¢
x 1 − x4 dx =
1 − u2 12 du =
R
1
(1
2
R
cos θ · cos θ dθ
"
where u = sin θ, du = cos θ dθ,
√
and 1 − u2 = cos θ
#
+ cos 2θ)dθ = 14 θ + 18 sin 2θ + C = 14 θ + 14 sin θ cos θ + C
√
√
= 14 sin−1 u + 14 u 1 − u2 + C = 14 sin−1 (x2 ) + 14 x2 1 − x4 + C
=
1
2
1
2
31. (a) Let x = a tan θ, where − π2 < θ <
Z
π
.
2
Then
√
x2 + a2 = a sec θ and
¯
¯√
Z
¯ x 2 + a2
x ¯¯
a sec2 θ dθ
¯
= sec θ dθ = ln|sec θ + tan θ| + C1 = ln¯
+ ¯ + C1
a sec θ
a
a
³
´
p
= ln x + x2 + a2 + C where C = C1 − ln |a|
dx
√
=
x2 + a2
Z
√
(b) Let x = a sinh t, so that dx = a cosh t dt and x2 + a2 = a cosh t. Then
Z
Z
dx
x
a cosh t dt
√
=
= t + C = sinh−1 + C.
a cosh t
a
x2 + a2
√
33. The average value of f (x) = x2 − 1/x on the interval [1, 7] is
"
#
Z 7√ 2
Z
where x = sec θ, dx = sec θ tan θ dθ,
1
x −1
1 α tan θ
dx =
· sec θ tan θ dθ
√
7−1 1
x
6 0 sec θ
x2 − 1 = tan θ, and α = sec−1 7
R
R
α
α
= 16 0 tan2 θ dθ = 16 0 (sec2 θ − 1) dθ
h
iα
= 16 tan θ − θ = 16 (tan α − α)
0
¡√
¢
= 16 48 − sec−1 7
35. Area of 4P OQ = 12 (r cos θ)(r sin θ) = 12 r 2 sin θ cos θ. Area of region P QR =
Let x = r cos u ⇒ dx = −r sin u du for θ ≤ u ≤ π2 . Then we obtain
Rr
r cos θ
√
r 2 − x2 dx.
R√
R
R
r2 − x2 dx = r sin u (−r sin u) du = −r 2 sin2 u du = − 12 r 2 (u − sin u cos u) + C
p
= − 12 r 2 cos−1 (x/r) + 12 x r2 − x2 + C
so
area of region P QR =
1
2
=
1
2
h
£
−r 2 cos−1 (x/r) + x
¡
2
p
r 2 − x2
0 − −r θ + r cos θ r sin θ
= 12 r 2 θ − 12 r 2 sin θ cos θ
¢¤
ir
r cos θ
and thus, (area of sector P OR) = (area of 4P OQ) + (area of region P QR) = 12 r 2 θ.
12 ■ TRIGONOMETRIC SUBSTITUTION
√
37. From the graph, it appears that the curve y = x2 4 − x2 and the line
√
y = 2 − x intersect at about x = 0.81 and x = 2, with x2 4 − x2 > 2 − x on
(0.81, 2). So the area bounded by the curve and the line is A ≈
√
R 2 £ 2√
¤2
¤
R2
£
x 4 − x2 − (2 − x) dx = 0.81 x2 4 − x2 dx − 2x − 12 x2 0.81 .
0.81
To evaluate the integral, we put x = 2 sin θ, where − π2 ≤ θ ≤
π
.
2
Then
dx = 2 cos θ dθ, x = 2 ⇒ θ = sin−1 1 = π2 , and x = 0.81 ⇒ θ = sin−1 0.405 ≈ 0.417. So
√
R2
R π/2
R π/2
R π/2
x2 4 − x2 dx ≈ 0.417 4 sin2 θ (2 cos θ)(2 cos θ dθ) = 4 0.417 sin2 2θ dθ = 4 0.417 12 (1 − cos 4θ) dθ
0.81
£
¤π/2
£¡
¢ ¡
¢¤
= 2 θ − 14 sin 4θ 0.417 = 2 π2 − 0 − 0.417 − 14 (0.995) ≈ 2.81
£¡
¢ ¡
¢¤
Thus, A ≈ 2.81 − 2 · 2 − 12 · 22 − 2 · 0.81 − 12 · 0.812 ≈ 2.10.
39. Let the equation of the large circle be x2 + y 2 = R2 . Then the equation of the small circle is x2 + (y − b)2 = r 2 ,
√
where b = R2 − r2 is the distance between the centers of the circles. The desired area is
√
√
√
R r £¡
¢ √
¤
Rr¡
¢
A = −r b + r2 − x2 − R2 − x2 dx = 2 0 b + r 2 − x2 − R2 − x2 dx
=2
Rr
0
b dx + 2
Rr√
Rr√
r 2 − x2 dx − 2 0 R2 − x2 dx
0
√
The first integral is just 2br = 2r R2 − r 2 . To evaluate the other two integrals, note that
R
R
R√
a2 − x2 dx = a2 cos2 θ dθ [x = a sin θ, dx = a cos θ dθ] = 12 a2 (1 + cos 2θ) dθ
¡
¢
= 12 a2 θ + 12 sin 2θ + C = 12 a2 (θ + sin θ cos θ) + C
³ x ´ a2 ³ x ´ √a2 − x2
³x´ xp
a2
a2
arcsin
+
+C =
arcsin
+
a2 − x2 + C
=
2
a
2 a
a
2
a
2
so the desired area is
h
ir h
ir
p
p
p
A = 2r R2 − r 2 + r 2 arcsin(x/r) + x r2 − x2 − R2 arcsin(x/R) + x R2 − x2
0
= 2r
p
R2 − r 2 + r
2¡ π ¢
2
h
− R2 arcsin(r/R) + r
p
0
i
R2 − r 2 = r
41. We use cylindrical shells and assume that R > r. x2 = r 2 − (y − R)2
p
g(y) = 2 r 2 − (y − R)2 and
V =
R R+r
R−r
2πy · 2
Rr
p
r2 − (y − R)2 dy =
Rr
−r
p
R2 − r 2 +
⇒
x=±
π 2
2r
p
− R2 arcsin(r/R)
r2 − (y − R)2 , so
√
4π(u + R) r2 − u2 du [where u = y − R]
√
Rr √
= 4π −r u r2 − u2 du + 4πR −r r 2 − u2 du
"
where u = r sin θ, du = r cos θ dθ
in the second integral
#
h ¡
¢3/2 ir
R
R π/2
2 π/2
= 4π − 13 r 2 − u2
+ 4πR −π/2 r2 cos2 θ dθ = − 4π
cos2 θ dθ
3 (0 − 0) + 4πRr
−π/2
= 2πRr
2 R π/2
−π/2
−r
£
(1 + cos 2θ) dθ = 2πRr 2 θ +
1
2
sin 2θ
Another method: Use washers instead of shells, so V = 8πR
the integral using y = r sin θ.
¤π/2
−π/2
Rrp
0
= 2π2 Rr2
r 2 − y2 dy as in Exercise 6.2.39(a), but evaluate