# Trigonometric Substitution Stewart Calculus

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```					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        x2     sa 2              a 2 sin 2             sa 2 1          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 t t by using the Substitution
Rule in reverse. To make our calculations simpler, we assume that t has an inverse func-
tion; 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

yf   x dx          yf       t t t t 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 restric-
tion 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     x2          x        a sin ,                                                                 1       sin 2       cos 2
2             2

sa 2     x2          x        a tan ,                                                                 1       tan 2       sec 2
2              2
3
sx 2     a2          x        a sec ,               0                 or                              sec 2           1   tan 2
2                       2

s9            x2
EXAMPLE 1 Evaluate         y         x   2            dx.

SOLUTION Let x           3 sin , where                         2                   2. Then dx           3 cos         d and

s9          x2       s9                9 sin 2           s9 cos 2           3 cos                3 cos

(Note that cos            0 because                       2                      2.) Thus, the Inverse Substitution Rule
gives
s9               x2                 3 cos
y            x   2        dx        y   9 sin 2
3 cos d

cos 2                       2
y   sin 2
d            y cot        d

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
3                         done either by using trigonometric identities to express cot in terms of sin         x 3 or
x        by drawing a diagram, as in Figure 1, where is interpreted as an angle of a right triangle.
¨                                      Since sin      x 3, 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
œ„„„„„
9-≈                           can simply read the value of cot from the ﬁgure:
FIGURE 1
s9               x2
x                                                                                              cot
sin ¨=                                                                                                                                 x
3

(Although        0 in the diagram, this expression for cot                                                   is valid even when          0.)
Since sin       x 3, we have      sin 1 x 3 and so

s9           x2                            s9         x2                          x
y         x   2
dx
x
sin   1
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                                         b
1                                                     or                y           sa 2           x2
b2                    a2                a   2
a
(a, 0)
0                         x   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       0 x a
FIGURE 2                                                                         a
≈   ¥                                                                                                               a       b
+ =1                                                                                           1
a@  b@                                        and so                                              4   A        y    0       a
sa 2             x 2 dx

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        x2       sa 2             a 2 sin 2                   sa 2 cos 2                   a cos               a cos

since 0                  2. Therefore

b        a                                            b             2
A     4
a
y  0
sa 2            x 2 dx              4
a
y   0
a cos         a cos d

2                                                2 1
4ab y                cos 2 d                       4ab y            2       1     cos 2       d
0                                             0

2ab     [            1
2   sin 2     ]   0
2
2ab
2
0       0

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

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

1
EXAMPLE 3 Find         y x sx   2           2         4
dx.

SOLUTION Let x             2 tan ,                           2                       2. Then dx               2 sec 2 d and

sx 2           4               s4 tan 2                    1        s4 sec 2              2 sec                 2 sec

Thus, we have
dx                         2 sec 2 d                          1       sec
y x sx   2           2         4
y   4 tan 2 2 sec                         4
y tan     2
d

To evaluate this trigonometric integral we put everything in terms of sin                                                                and cos :

sec                 1               cos 2           cos
tan 2              cos               sin 2           sin 2

Therefore, making the substitution u                                            sin , we have

dx                               1        cos                  1       du
y           2
x sx 2                   4          4
y    sin 2
d
4
y   u2
1            1                           1
≈+4
œ„„„„„
C                               C
4            u                        4 sin
x
csc
¨
C
4
2
We use Figure 3 to determine that csc                                                sx 2         4 x and so
FIGUR E 3
x                                                                                   dx                       sx 2 4
tan ¨=
2                                                                     y x sx  2     2        4                 4x
C

x
EXAMPLE 4 Find         y sx         2            4
dx.

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
x                          1            du
y    sx      2            4
dx
2
y su               su         C       sx 2          4       C

NOTE  Example 4 illustrates the fact that even when trigonometric substitutions are pos-
■

sible, they may not give the easiest solution. You should look for a simpler method ﬁrst.

dx
EXAMPLE 5 Evaluate              y sx             2           a2
, where a               0.

SOLUTION We let x     a sec , where 0                                                          2 or                     3       2. Then
dx      a sec    tan d and

sx 2        a2                  sa 2 sec 2                  1            sa 2 tan 2           a tan                 a tan

Therefore
dx                          a sec tan
y sx             2        a   2         y       a tan
d

y sec         d          ln sec                tan            C
4 ■ TRIGONOMETRIC SUBSTITUTION

The triangle in Figure 4 gives tan                                         sx 2                     a 2 a, so we have
x
≈-a@
œ„„„„„                                                 dx                               x                    sx 2             a2
y sx       2         a   2
ln
a                             a
C
¨
a                                                                                              ln x                     sx 2             a2             ln a                  C
FIGU RE 4
x                           Writing C1        C             ln a, we have
sec ¨=
a
dx
y sx        2           a2
ln x                     sx 2              a2             C1

3 s3 2               x3
EXAMPLE 6 Find            y
0             4x 2            9       3 2    dx.

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 substitu-
3
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                           3s3 2, tan                               s3, so                       3.

27
3 s3 2                 x3                                         3        8tan3                3
y                      2              3 2      dx            y                                      2   sec 2 d
0            4x              9                            0               27 sec3

3               3   tan 3                        3            3   sin3
16   y   0           sec
d                     16   y   0        cos2
d

3               3   1         cos 2
16   y   0                   cos 2
sin d

Now we substitute u                     cos         so that du                                  sin d . When                                      0, u        1; when
3, u 1.2
Therefore

3 s3 2                x3                                   3       1 2         1               u2                  3       1 2                   2
y0               4x   2
9    3 2   dx                  16   y  1                    u   2        du            16   y1
1       u       du
1 2
1
3
16       u                                      3
16   [( 1
2           2)               1       1   ]     3
32
u           1

x
EXAMPLE 7 Evaluate                 y s3             2x                x2
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           x2          3             x2           2x                      3         1             x2            2x          1
2
4             x            1

This suggests that we make the substitution u                                                           x        1. Then du                           dx and x         u    1, so

x                                                u          1
y s3                2x            x2
dx                  y    s4          u2
du
TRIGONOMETRIC SUBSTITUTION ■ 5

We now substitute u                      2 sin , giving du                      2 cos d and s4                                     u2                   2 cos , so

x                                2 sin                   1
y s3   2x      x2
dx          y         2 cos
2 cos d

y       2 sin                   1 d

2 cos                                  C
u
s4              u2               sin      1                   C
2
x            1
s3             2x                x2           sin   1                                           C
2

Exercises
s1         x2                                                                            t
x                                                                                               t2
1–3     Evaluate the integral using the indicated trigonometric                                                                                     2 3                                                                         1
substitution. Sketch and label the associated right triangle.                                                                             21.   y         x 3s4          9x 2 dx                                  22.       y       sx 2                     1 dx
0                                                                        0

1
1.    yx       2                           dx ;       x        3 sec                                                                                                                                                                              dt
sx 2              9                                                                                                23.   y s5              4x          x 2 dx                              24.       y st             2        6t               13

2.    yx       3
s9        x 2 dx ;                 x    3 sin                                                                                               1                                                                             x2
25.   y s9x         2
dx                      26.       y s4x                                 dx
6x         8                                                                     x2
x3
3.    y sx                           dx ;         x       3 tan                                                                                                dx                                                                                 dx
2        9                                                                                                        27.   y                                                                 28.       y
x2         2x         2   2
5               4x x 2                5 2
■             ■             ■            ■            ■        ■       ■         ■             ■            ■         ■      ■   ■

4–30                   Evaluate the integral.                                                                                                                                                                                        2               cos t
29.   y x s1              x 4 dx                                        30.       y                                   dt
2 s3        x      3                                                                                                                                                                                          0                s1         sin 2 t
4.    y                                       dx
0           s16                x2                                                                                              ■         ■         ■          ■            ■             ■        ■          ■                ■               ■             ■         ■   ■

2          1                                                               2                                                31. (a) Use trigonometric substitution to show that
5.    y                                      dt                         6.       y       x 3 sx 2            4 dx
s2      t 3 st 2              1                                           0
dx
1                                                                    sx 2 a 2                                                              y sx       2
ln ( x        sx 2                   a2)              C
7.                                                                                                                                                                                        a2
y       x 2 s25                x   2
dx                         8.       y           x4
dx

dx                                                                         t5                                           (b) Use the hyperbolic substitution x                                                a sinh t to show that
9.    y sx          2
10.       y st          2
dt
16                                                                          2
dx                                 x
11.
1                                                                                   y sx         2        a2
sinh       1
a
C
y s1                  4x dx2
12.       y
0
x sx      2        4 dx

sx 2 9                                                                               du                                               These formulas are connected by Formula 3.9.3.
13.       y              dx                                                14.       y u s5
x3                                                                                         u2
32. Evaluate
x2                                                                     dx                                                                                                       x2
15.       y                                      dx                        16.       y
a2         x2          3 2
x 2 s16x 2               9                                                                       y     x   2
a2       3 2
dx

x                                                                                  dx
17.       y sx                           dx                                18.       y                 2
(a) by trigonometric substitution.
2
7                                                            ax                b2    3 2
(b) by the hyperbolic substitution x                                                a sinh t.
6 ■ TRIGONOMETRIC SUBSTITUTION

33. Find the average value of f x                   sx 2          1 x, 1     x   7.       where is the charge density per unit length on the rod and 0
is the free space permittivity (see the ﬁgure). Evaluate the inte-
34. Find the area of the region bounded by the hyperbola                                  gral to determine an expression for the electric ﬁeld E P .
9x 2   4y 2       36 and the line x             3.
y
1
35. Prove the formula A       r 2 for the area of a sector of a circle
2
P (a, b)
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                                           0                              L   x
triangle POQ and the area of the region PQR in the ﬁgure.]

y
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                                                    r

; 36. Evaluate the integral
R
dx
y    x 4 sx 2        2

Graph the integrand and its indeﬁnite integral on the same
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 verti-
; 37. Use a graph to approximate the roots of the equation                                   cal. If the depth of the water is 7 ft, what percentage of the
x 2 s4 x 2 2 x. Then approximate the area bounded by                                   total capacity is being used?
the curve y x 2 s4 x 2 and the line y 2 x.
41. A torus is generated by rotating the circle x 2 y R2            r2
38. A charged rod of length L produces an electric ﬁeld at point                          about the x-axis. Find the volume enclosed by the torus.
P a, b given by

L a                  b
EP           y   a     4    0    x2        b2   3 2
dx
TRIGONOMETRIC SUBSTITUTION ■ 7

19. ln (s1                                   1) x                                                           64
x2                                 s1           x2       C         21.
1215

23.   2   sin           x            2 3                  2   x         2 s5             4x       x2       C
1                                                     1
1. sx 2       9 9x      C      3.    3   x2        18 sx 2    9       C              25.   3   ln 3x                 1            s9x      2            6x           8        C
1             1                                                       2
5.       24    s3 8
1
7.         s25        x 2 25x       C                  27.   2   tan           x            1            x            1        x           2x        2        C
4
1                             1
9. ln (sx 2   16 x) C          11. sin 1 2x
1
x s1         1
4x 2   C   29.   4   sin 1 x 2                 4   x 2 s1                x4        C
4              2
1      1
13. 6 sec x 3        sx 2  9 2x  2
C                                               33.
1
6   (s48              sec          1
7)               37. 0.81, 2; 2.10
15. (x sa 2    x 2 ) sin 1 x a     C    17. sx 2    7                     C          39. r sR        2           r   2             2
r 2               R 2 arcsin r R                     41. 2        2
Rr 2
8 ■ TRIGONOMETRIC SUBSTITUTION

Solutions: Trigonometric Substitution

π                 3π
1. Let x = 3 sec θ, where 0 ≤ θ <                 2   or π ≤ θ <     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                            Z                                                                                                             √
1                                  1                                              1
R                1                       1    x2 − 9
√       dx =                                    3 sec θ tan θ dθ =                  9
cos θ dθ =   9
sin θ + C =                     +C
x2 x2 − 9                      9 sec2 θ · 3 tan θ                                                                                  9     x

3π
Note that − sec(θ + π) = sec θ, so the ﬁgure is sufﬁcient for the case π ≤ θ <                                       2
.

3. Let x = 3 tan θ, where − π < θ <
2
π
2.   Then dx = 3 sec2 θ dθ and
p               p                              q                      √
x2 + 9 =        9 tan2 θ + 9 =                 9(tan2 θ + 1) =     9 sec2 θ

= 3 |sec θ| = 3 sec θ for the relevant values of θ.

Z                        Z                            Z                   Z
x3                  33 tan3 θ
√       dx =                     3 sec2 θ dθ = 33 tan3 θ sec θdθ = 33 tan2 θ tan θ sec θ dθ
x2 + 9                 3 sec θ
3R ¡             ¢                   R¡ 2    ¢
=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
3¡ 1 3      ¢           3¡ 1   3       ¢         3 1 x +9              x2 + 9
= 3 3 u − u + C = 3 3 sec θ − sec θ + C = 3                           −             +C
3     33             3
¡            ¢3/2          p                                 ¡          ¢p
=   1
3
x2 + 9            −9       x2 + 9 + C     or         1
3
x2 − 18   x2 + 9 + C
√                   π                                         π
5. Let t = sec θ, so dt = sec θ tan θ dθ, t =                   2   ⇒ θ=           4
,   and t = 2 ⇒ θ =                     3
.   Then

Z   2                             Z   π/3                         Z π/3              Z π/3
1                          1                                   1
√
√       dt =                         sec θ tan θ dθ =                dθ =       cos2 θ dθ
2   t3 t2 − 1              π/4         sec3
θ tan θ                     π/4  sec 2θ
π/4
R π/3 1                    1
£    1
¤π/3
= π/4 2 (1 + cos 2θ) dθ = 2 θ + 2 sin 2θ π/4
h³      √ ´     ¡            ¢i     ³      √       ´        √
= 1 π + 1 23 − π + 1 · 1 = 1 12 + 43 − 1 = 24 + 83 − 1
2     3 2           4    2          2
π
2
π
4

7. Let x = 5 sin θ, so dx = 5 cos θ dθ. Then
Z                      Z
1                        1
√          dx =                          5 cos θ dθ
x2 25 − x2              52 sin2 θ · 5 cos θ
R
= 25 csc2 θ dθ = − 25 cot θ + C
1                     1

√
1 25 − x2
=−                   +C
25      x
TRIGONOMETRIC SUBSTITUTION ■ 9

9. Let x = 4 tan θ, where − π < θ <
2
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               Z
dx              4 sec2 θ dθ
√        =                    = sec θ dθ = ln |sec θ + tan θ| + C1
x2 + 16            4 sec θ
¯√              ¯
¯ x2 + 16     x¯           ¯√              ¯
= ln ¯
¯            + ¯ + C1 = ln ¯ x2 + 16 + x¯ − ln |4| + C1
4       4¯
¡√             ¢
= ln x2 + 16 + x + C, where C = C1 − ln 4.
√
(Since    x2 + 16 + x > 0, we don’t need the absolute value.)

11. Let 2x = sin θ, where − π ≤ θ ≤
2
π
2
.   Then x =       1
2
sin θ,
√                           q
dx = 1 cos θ dθ, and 1 − 4x2 =
2                                               1 − (2x)2 = cos θ.
R√             R     ¡        ¢      R
1 − 4x2 dx = cos θ 1 cos θ dθ = 1 (1 + cos 2θ) dθ
2           4
¡           ¢
= 1 θ + 1 sin 2θ + C = 1 (θ + sin θ cos θ) + C
4     2               4
h              p         i
= 1 sin−1 (2x) + 2x 1 − 4x2 + C
4

13. Let x = 3 sec θ, where 0 ≤ θ < π or π ≤ θ < 3π . Then
2               2
√
dx = 3 sec θ tan θ dθ and x  2 − 9 = 3 tan θ, so

Z √ 2              Z                                Z
x −9              3 tan θ                    1   tan2 θ
3
dx =           3θ
3 sec θ tan θ dθ =            dθ
x               27 sec                      3   sec2 θ
R                         R
=    1
3       sin2 θ dθ =       1
3
1
2
(1   − cos 2θ) dθ = 1 θ −
6
1
12
sin 2θ + C = 1 θ −
6
1
6
sin θ cos θ + C

³x´ 1                 √                     ³ x ´ √x2 − 9
1                              x2 − 9 3      1
=      sec−1    −                            + C = sec−1      −        +C
6        3   6                  x     x      6        3    2x2
15. Let x = a sin θ, where − π ≤ θ ≤
2
π
2.   Then dx = a cos θ dθ and
Z                             Z                                    Z
x2 dx                    a2 sin2 θ a cos θ dθ
=                            =               tan2 θ dθ
(a2 − x2 )3/2                     a3 cos3 θ
Z
¡             ¢
=           sec2 θ − 1 dθ = tan θ − θ + C

x            x
=√        − sin−1 + C
a2 − x2        a

Z                           Z                                p
x          1              1                   √
17. Let u = x2 − 7, so du = 2x dx. Then                      √       dx =               √ du =      1
2
·2    u + C = x2 − 7 + C.
x2 − 7      2               u
10 ■ TRIGONOMETRIC SUBSTITUTION

19. Let x = tan θ, where − π < θ < π . Then dx = sec2 θ dθ
2         2
√
and 1 + x  2 = 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¯           2          ¯         2     ¯ √
¯
= ln ¯          − ¯  ¯ + 1 + x + C = ln ¯ 1 + x − 1 ¯ + 1 + x2 + C
x       x         1             ¯       x       ¯

21. Let u = 4 − 9x2           ⇒    du = −18x dx. Then x2 =                   1
9
(4 − u) and
R 2/3       √            R0 1            ¡ 1¢                                           R4³                  ´
0
x3 4 − 9x2 dx = 4 9 (4 − u)u1/2 − 18 du =                                1
162       0
4u1/2 − u3/2 du
h                      i4             £ 64            ¤
8 3/2
=    1
162       3u      − 2 u5/2
5             =    1
162     3    −   64
5        =    64
1215
0

Or: Let 3x = 2 sin θ, where − π ≤ θ ≤
2
π
2
.

23. 5 + 4x − x2 = −(x2 − 4x + 4) + 9 = −(x − 2)2 + 9. Let
x − 2 = 3 sin θ, − π ≤ θ ≤ π , so dx = 3 cos θ dθ. Then
2        2
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                       ¡            ¢
= 9 (1 + cos 2θ) dθ = 9 θ + 1 sin 2θ + C
2                       2      2

= 9θ +
2
9
4
sin 2θ + C = 9 θ + 9 (2 sin θ cos θ) + C
2     4
µ       ¶                √
9       −1   x−2       9 x−2          5 + 4x − x2
=       sin              + ·          ·                 +C
2               3      2     3             3
µ       ¶
9            x−2       1         √
=       sin−1            + (x − 2) 5 + 4x − x2 + C
2               3      2
Z    Z    1
dx                du
25. 9x2 + 6x − 8 = (3x + 1)2 − 9, so let u = 3x + 1, du = 3dx. Then                      √               =   √3        . Now
9x 2 + 6x − 8       u2 − 9
π
√
let u = 3 sec θ, where 0 ≤ θ <              2
or π ≤ θ < 3π . Then du = 3 sec θ tan θ dθ and u2 − 9 = 3 tan θ, so
2
Z                   Z                                                                ¯    √                                          ¯
1
du                sec θ tan θ dθ      R                                       ¯ u + u2 − 9                                    ¯
√3      =                           = 1 sec θdθ = 1 ln|sec θ + tan θ| + C1 = 1 ln¯
3          3                          3   ¯
¯ + C1
¯
u2 − 9                 3 tan θ                                                        3
¯     p        ¯         ¯          p               ¯
1     ¯              ¯         ¯                          ¯
=    3
ln¯u + u2 − 9¯ + C = 1 ln¯3x + 1 + 9x2 + 6x − 8 ¯ + C
3

27. x2 + 2x + 2 = (x + 1)2 + 1. Let u = x + 1, du = dx. Then
Z                               Z                         Z                        "                                       #
dx                             du                  sec2 θdθ                where u = tan θ, du = sec2 θ dθ,
=                            =
(x2 + 2x + 2)2                    (u2 + 1)2               sec4 θ                       and u2 + 1 = sec2 θ
R             R
=   cos2 θ dθ = 1 (1 + cos 2θ) dθ = 1 (θ + sin θ cos θ) + C
2                     2
·                  ¸         ·                            ¸
1      −1        u           1       −1              x+1
=     tan u +            +C =      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√      ¡    ¢                   R                             where u = sin θ, du = cos θ dθ,
x 1 − x4 dx =   1 − u2 1 du =
2
1
2
cos θ · cos θ dθ                  √
and 1 − u2 = cos θ
1
R   1
=   2
+ cos 2θ)dθ = 1 θ + 1 sin 2θ + C = 1 θ + 1 sin θ cos θ + C
2
(1        4     8              4     4
√                                 √
= 1 sin−1 u + 1 u 1 − u2 + C = 1 sin−1 (x2 ) + 1 x2 1 − x4 + C
4           4                  4              4

√
31. (a) Let x = a tan θ, where − π < θ <
2
π
2
.   Then             x2 + a2 = a sec θ and

Z                   Z             Z                                      ¯√           ¯
dx             a sec2 θ dθ                                         ¯ x 2 + a2  x¯
√        =                                                           ¯
= sec θ dθ = ln|sec θ + tan θ| + C1 = ln¯          + ¯ + C1
x2 + a2           a sec θ                                                a      a¯
³     p         ´
= ln x + x2 + a2 + C where C = C1 − ln |a|

√
(b) Let x = a sinh t, so that dx = a cosh t dt and x2 + a2 = a cosh t. Then
Z                Z
dx             a cosh t dt                  x
√         =                   = t + C = sinh−1 + C.
x2 + a2          a cosh t                    a

√
33. The average value of f (x) = x2 − 1/x on the interval [1, 7] is
Z 7√ 2                  Z                         "                                       #
1          x −1          1 α tan θ                      where x = sec θ, dx = sec θ tan θ dθ,
dx =              · sec θ tan θ dθ    √
7−1 1         x            6 0 sec θ                         x2 − 1 = tan θ, and α = sec−1 7
Rα                 Rα
= 1 0 tan2 θ dθ = 1 0 (sec2 θ − 1) dθ
6                 6
h        iα
1
= 6 tan θ − θ = 1 (tan α − α)
6
0
¡√              ¢
= 1 48 − sec−1 7
6

Rr            √
35. Area of 4P OQ = 1 (r cos θ)(r sin θ) = 1 r 2 sin θ cos θ. Area of region P QR =
2                     2                                                                 r cos θ
r 2 − x2 dx.
Let x = r cos u ⇒ dx = −r sin u du for θ ≤ u ≤ π . Then we obtain
2

R√             R                           R
r2 − x2 dx = r sin u (−r sin u) du = −r 2 sin2 u du = − 1 r 2 (u − sin u cos u) + C
2

−1
p
1 2               1    2 − x2 + C
= − 2 r cos (x/r) + 2 x r

so

h                          p              ir
area of region P QR =         1
2       −r 2 cos−1 (x/r) + x       r 2 − x2
r cos θ
1
£       ¡   2                     ¢¤
=       2
0 − −r θ + r cos θ r sin θ

= 1 r 2 θ − 1 r 2 sin θ cos θ
2         2

and thus, (area of sector P OR) = (area of 4P OQ) + (area of region P QR) = 1 r 2 θ.
2
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√                     ¤       R2      √               £    ¤2
0.81
x 4 − x2 − (2 − x) dx = 0.81 x2 4 − x2 dx − 2x − 1 x2 0.81 .
2

To evaluate the integral, we put x = 2 sin θ, where − π ≤ θ ≤
2
π
2
.   Then

dx = 2 cos θ dθ, x = 2 ⇒ θ = sin−1 1 = π , and x = 0.81 ⇒ θ = sin−1 0.405 ≈ 0.417. So
2
R2      √              R π/2                                  R π/2               R π/2
0.81
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 1 (1 − cos 4θ) dθ
2
£            ¤π/2      £¡ π     ¢ ¡                   ¢¤
= 2 θ − 1 sin 4θ 0.417 = 2 2 − 0 − 0.417 − 1 (0.995) ≈ 2.81
4                                      4
£¡        1   2
¢ ¡            1      2
¢¤
Thus, A ≈ 2.81 − 2 · 2 − 2 · 2 − 2 · 0.81 − 2 · 0.81          ≈ 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
Rr                Rr√               Rr√
=2     0
b dx + 2      0
r 2 − x2 dx − 2 0 R2 − x2 dx
√
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] = 1 a2 (1 + cos 2θ) dθ
2
¡             ¢
= 1 a2 θ + 1 sin 2θ + C = 1 a2 (θ + sin θ cos θ) + C
2        2               2
√
³ x ´ a2 ³ x ´ a2 − x 2                       ³x´ xp
a2                                           a2
=     arcsin      +                     +C =        arcsin       +    a2 − x2 + C
2          a      2 a        a               2            a     2
so the desired area is
p             h             p       ir h                 p        ir
A = 2r R2 − r 2 + r 2 arcsin(x/r) + x r2 − x2 − R2 arcsin(x/R) + x R2 − x2
0                                               0
p                               h                    p              i       p
2¡ π ¢
= 2r       R2 − r 2 + r        2     − R2 arcsin(r/R) + r        R2 − r 2 = r          R2 − r 2 +   π 2
2r    − R2 arcsin(r/R)
p
41. We use cylindrical shells and assume that R > r. x2 = r 2 − (y − R)2                   ⇒     x=±        r2 − (y − R)2 , so
p
g(y) = 2 r 2 − (y − R)2 and
R R+r               p                        Rr            √
V =      R−r
2πy · 2        r2 − (y − R)2 dy =   −r
4π(u + R) r2 − u2 du [where u = y − R]
"                                            #
Rr √                Rr √                                    where u = r sin θ, du = r cos θ dθ
= 4π −r u r2 − u2 du + 4πR −r r 2 − u2 du
in the second integral
h ¡           ¢3/2 ir      R π/2                                     R π/2
= 4π − 1 r 2 − u2
3                  + 4πR −π/2 r2 cos2 θ dθ = − 4π (0 − 0) + 4πRr 2 −π/2 cos2 θ dθ
3
−r

2 R π/2                           £                     ¤π/2
= 2πRr       −π/2
(1 + cos 2θ) dθ = 2πRr 2 θ +     1
2   sin 2θ    −π/2
= 2π2 Rr2
Rrp
Another method: Use washers instead of shells, so V = 8πR               0
r 2 − y2 dy as in Exercise 6.2.39(a), but evaluate
the integral using y = r sin θ.

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