1 1)
1  2
cos|+ 
form − ln( = +
+ln
.||2+  =cos
2 +
+
ln |cos
√
2 =
csc  − 22tan
+4csc
 
 2
 −
√
sin 2 
sin

1

2
2
2  = √  ⇒
√

=
18.
Let

=
.
Then
.(2
)
= 2  = 2(
14. 
Let

=
ln(1
+

),

=

⇒

=
,

=
Then


 − ).
−1

√
√


2
2
−1 22 
−1
1
2
2

1
+

1
1
2 =6  sin  
6
+ 3) = ,
=
28. sin 2
  = sin
 · 2(
 −
=3,, 2  =  ]
s
   1[
=

=
2
ln
||
−
+
5.

=

+ ln |csc  
− cot | − ln |sin |
7.2.39]√
= − cot +2

2
2 [by Exercise
 +
√
√
0 ( − 3)
sin  2 =   22 −3
−3
(2 2+ 1)−
1
2 −3+
2
2 
2
 = =

[−
cos

+
sin
]
+

[integration
by
parts]
−

cos

+
sin2 
+
ln(1
+

)

=

ln(1
+

)
−

ln(1
+

)
−
2
19. Let  =  . =
Then


=



=


=

+

=

+
.



2
1
+

1
+

= (2ln 1 + 6)
−
(2
ln
3
+
2)
=
4
−
2
ln
3
or
4
−
ln
9
√1 − cos  √


cos) + .
The answer can
be written
− ln(1
= −2
cosas  +2 2 sin
 ++
20. Since 2 is a constant,  2  =sin
 + .2
−1
− 1
1
ln(1+ 2 ) − 2
)
−
2
=

ln(1
+

2 1= −1
 
1
1 + 2 tan2  + 
1+
√
√
6. Let  = 2 . Then  = 2  ⇒
=
= sin √ +  = sin−1 √ + .
√
√
√

3 −24   2
3 − 2  2
3
3 1 2
2
29.
2=
ln
+
− 1 arctan
,3  = 2
4
1 3
21. Use
Let
integration
=
, .
so by
thatparts

=with
sec
and
2
 =
. Then

⇒
 (21 
)
4 = .1Now use parts with
37.
Let

=
tan
Then
=


⇒
tan

sec
== 0 arctan
  =
= 4.
15. Let  = ,  = sec  tan   ⇒  =0 ,  = sec . Then
4

0
1


  4
1arctan2   √2 −4

 
1

−4
1
1
+


1
7. Let

=
arctan
.
Then

=
⇒

=


=

4 −
.
,

=

.
Thus,

=
arctan
,

=
2

⇒

=
sec  tan
 −1 +sec
==2 sec
−ln
√   = 1 sec
√1+
2 |sec √+ tan | + .
=
= . Then
+√
 = √−4 = ,
 2
1
+

2
2
2
2
2
−1
−4
+  −1
 − 1 4 
2+tan  − 1
 −1
 −1
38. The integrand
is an odd
function,so
 Theorem 5.5.7(b)].
 4  = 0 [by
 1+
 

 
2 , so that

16. Let
cos

1
−4
 −21 = 2 .2 Then
 21 + 
−=
2
21
2
2 −
ln  arctan
+ 
 =2 ln=
+−
1 − 1⇒−√  − 1

1
−  +
− −1
1 +
.

=

ln

+
−
arctan

−
+
arctan


=

= − 1
= +arctan
=
(
+
1)
+
(
−
5).
Setting
2 =
gives
8. 2 =



1 +2+ 1)
12+−
21
 − 4
( − 5)(
 − 5  −
+ 11 

3− 5
1
1
2
12
−12
32
12
1



√
√
(

 =
( ) =
 =
−
)  =
− 2
+
 12 2 sec  tan
1
2 3 1
1
√

√. Setting
√
1 + 2so√
1 + 2 =√5 gives 4 =

6,
 = 23 2
. Now
 tan+
 (so

=  +2
 =
 = . Now
39. −2
Let==−6,
=
sec ,
so−=
that
=if
sec
.
−1 3 
+
arctan
 Then
or
+
1)
arctan

−
arctan
2

−
1
≥
0
−
1
if

≥
0

  − sec  √
sec
 − 
( − 1)
1
30. | − 1| =
(1 + 2 )32 − (1 + 2=
+   or 13 (2 −
=
 2) 1+ 2 + 
)12
4
4
if 13
0
−( 3− 41) if −1 − 1  0
1−

23
2
1


1

=
+
ln
|
−
5|
+
ln
|
+
1|

=
ln  −101)
=
+
1=
=11 to get 1 =3 . Set  = 03to get 1 = −,
so  = −1.
2(

 +.
  20 12 −⇒
4
− 5 Set
0
+
21 + 
Then
22. Let
−
=5  = .
(
,−so
1(1that

2
17.
−
cos
  (ln
= )
+ cos
2)
  + 12 0  cos 2 0
=1)
2 0
2 20 
0
0
2

 


Thus, −1 |
− 1|  = −1
(1 −  ) =+2 ln
( −1 1)
 5−− 1 ln 1+= − 1−ln5
 
0 1+
ln 
5 −=23 ln


 3 −1
0
3 √ 3
3

−1 11   1


ln

1

=
,
 =
cos 2 
 |
2−
1+
1 =
1 ln
Thus,
=
+2 = + 1 
ln ||
1|
+
 1=+ln(ln
|sec
−
√
2+

=
)
.
−
=
−1
2− 
2+ 1|
−11ln |sec | +  [or ln |1 − cos | + ].
= 122 12

sin
2
−
sin
2


−
1
2
2
2
2
0 ( − 2) − (1 − 0)
)+
 += 2 −
0 2
 =
= ,
sin32
 (−1
+0(0
22 − 1) −
 −

 + 21 + (ln )  =
=
=
+
. Multiply by ( + 4)( − 1) to get  + 2 = ( − 1) + ( + 4).
9. 2
 + 3 − 4
(
+ 4)( −11)
1
 1  + 4  −
1 2
1 2
1 2
− 2 cos7.2,
06
−in+
2 0sin=6
 + 18 (1
1 1) = 4 
4 Section
4
31.
in 2Example
5, 4  +2(a)
22
−
3
−2 =
40. As
Using
product
formula
[sin(6 − 3)SECTION
+ sin(6
+ STRATEGY
3)] = 12 (sin
3 + sin 9). Thus,
¤ 677
7.5+
FOR
3  cos 3 =
2 −4⇒
=
3
+
=
3
+
+
+ 22−2
==
(
2) ⇔
+ (
− 254).
Setting
23. Substituting
1
for

gives
3
=
5
⇔

=
.
Substituting
for
6gives
−5
=
.INTEGRATION
Thus,
5
2

−
2
−
8
(
−
4)(
+
2)

−
4

+
2



c
√
√
° 2016 Cengage
May not
to a publicly
in whole
or in part.

 or posted
 accessible1 website,

be1scanned, copied,or duplicated,
  All Rights Reserved.


 4  Learning.
4 −1
4 =
1 +35
 + sin 9)
 12 − 13 cos33 −
9 cos9
1+
1 +3
sin
cos
(sin
3
47. Let 1=+−
1, =
so that√
=6
.
Then
25

2 =
0+ 2
0 2√
23
5 √
√
√
·

=

=
+
=
sin

− 1 − 2 + .
0
 = 41 gives
so

=
.
Setting

=
−2
gives
10
=
−6,
so

=
−
.
Now

=
+

=
ln
|
+
4|
+
ln
|
−
1|
2
2
2
−  46 =6,
3
3
1
−

1
+

1
−

1
−

1
−

2 + 3 − 4
4
1−
5
 1 1


5
2
  3 −4 2 =
 1+

1
−4
31 + 2
− − − −4
= 12 49(
+−149 +=349−2 + 32−3 + −4 ) 
 3 ( −
( + 1) 

=

=
(

2 +
3 39 +3 +31)
9  =
2 1)

2
3 − 2
233 2 ln53
5 2) + 3 ln 3 − 2 (ln 2 + ln 3)
8 +−35 ).
ln
6 ln
+ |
0 −
=4|25 (3
(1 +5−)(1
Another
method: 
Substitute
= =
=
3 +
− ln
ln | +5 2| + .5
3 =−3 5+ln23
3
3
−2+ 2 1 −3
2 − 2 − 8 = ln || − 3−1
−−
4 3 
−

+

=
ln
|
−
1|
−
3(
− 1)−1 − 32 ( − 1)−2 − 13 ( − 1)−3 + 
2
2= 4 2ln 2 + 1 3ln 3, or 1 ln 48
41. Let  = ,  = tan   = sec  −
1  5 ⇒  5=  and  = tan  − . So

5
 3 3
 1
 
2



+
= 3,
√ ) sec  2= 1 (1 + 22 tan 
√
 √
24. (1 + tan
tan2 )
 sec     
1
2 , so
−
32.

=


23 1
2 2  = 0
2
48. Let =
1
−


=
−

,
and
2

=
−2
. Then
  2tan
− −
1−
21−2 (− ).
2 1

tan


=
(tan

−
)
−
(tan
−
)

=
cos(1)
1
1

=
−3

0
1 + 2 + 
 − ln |sec |
 ⇒  = − ,  = −2sin
.
Then
10. Let1  = ,  =3
3
2
2

−
1)
sec
]

=
(2
sec

tan

+
) 
=
[sec

+
2
sec

tan

+
(sec
¤ 677
 1
SECTION 7.5 STRATEGY sec
FOR INTEGRATION
 √
1 2

2

3
3
1
1
1
=

tan

−

−
ln
|sec
|
+

Now
let

=
2
−
,
so

=
2
−
,
and
2

=
−.
Thus,

=
−
=
−
(
−

)
=
(
−
)
2








1
3
3
3 
|sec1  + tan
1  tan
1 3= 2√1sec  + 2 (sec
cos(1)
1 |) +  1[by Example 7.2.8]
1  + ln
 = 2. −
 √
 0  =  −
47. Let
1, so
that 
Then 2 sin
sin

=2 − sin
− cos
+ .√
=−
√3
2






2
2
4
4 3
2 5


2
−

(−
)
=
(2
−

)
(2
)
=
(4
−
2
)

=

−




33. Using
product
formula
2(c)
in
Section
7.2,

1
1
1
3
5 1 1
 (12


 1 31 + 12 −4
+
4)
−
3
3
3
−4
3
2
−4
−1
−2
−3
−4
1
+ 1)  
31 + 3 +
1)
(+ 1|+ 3= (4+−3
= = (
 =
=  ( +
25.  ( − 1)  
ln 4)+
−(0 −) 
0) = 4 − ln 4
4−
= 4−=ln |3
√
√
√


1 Learning.
1
1
c
°
2016
Cengage
All
Rights
Reserved.
May
not
be
scanned,
copied,
or
duplicated,
or
posted
to
a
publicly
accessible
website,
in
whole
or
in part.
1
+
3
3
+
1
3
+
1
0
8 Reserved.
14 + cos
0 Learning.
+
cos(2May
+04not
6)]
= 16
[cos(−4)
8] to=a publicly
4 + cos
8).
Thus,
cos0 2 cos°
c 6
2016=
Cengage
be2scanned,
copied, or duplicated,
or posted
accessible
website,
in whole
or in part.
= 83−All6)
2Rights
−−1
2 [cos(2
2 (cos
−2 3 −1 5 −3=2 15 2 − 15
−1
−2
3
1
5 23 −
= ln || − 3 − 2  − 3   +  = ln | − 1| −
1) − 3 ( − 1)−3 + 

 3( − 1)1 − 2 ( −
1
1
1
1
1
2 4 + cos 8)  =
cos3
22 √
cos16  = 2 3
(cos
sin 4 + 8 sin 8 +  = 8 sin
4 + 16 sin 8 + .

+
+1
2 
4 +
1
+ =24  . ⇒
Multiply
(
+ 1)(
26. Let
get
49.
 = 4 + 1 =⇒ 2 2 = 4 + 1 =⇒ 2 
.
So 2 + 1)to0 √
√
 1=by2 
3 + 2 √
2
2

+

+
1
(
+
1)(
+
1)

+
1

+
1
48. Let  = 1 − 2 , so  = 1 −  , and 2  = −2 . Then 0  2 − 1 − 2  = 1 2 −  (− ).




 2
1
2
√
 + ( + ). Substituting −1 for 
 12 + (
 ⇔ 32 + 1 =( + )
−
1)
1 + (
3
+
1
=
(
+
1)
+
)(
+
1)
+
2
2
+

Now let  = 2 −√, so  =
2  = −.
[by Formula 19]
2=− , and
= 2 2 ln
= 2 Thus,
1
 accessible website,
2or−
2
c 2016 Cengage
°
Learning.
May
not
bescanned, copied,
duplicated,
or posted to 
a publicly
in whole or in part.

1
+
1
(
−
1)
 4
+ 1All Rights Reserved.
4
2
√ gives 3 =  +  = 2 + 

 0 4 = 2 ⇔  = 2.√2Equating
⇔

=
1.
Equating coefficients of 
gives
coefficients
of

√
2


√
√
 2

2
1 −=1 
2 −  (− ) =
(2 − √
)4
(2+)
(4 2 − 2 4 )  = 43  3 − 25  5 1

gives
0 =  +  = 1 + 1 ⇔= ln
= −1. Thus,  +
1
4 + 1 + 1 1
√
√
√







 1
8
8
4
2
16 
14
2
− 15
1 2 − 15 1

1
32 +=1 3 2 − 5 21 − 23 − 5 =
=
 +
 −

=

+

1
3
2 +1
2 +1
 

2 +  + √
1
+1

2 + 1

0√  +
0 2 +1
√
=
50. As in Exercise
49, let  = 2 4 + 1.0Then
.
Now
=
8


2
1 =
2 1
 1 ⇒ 2
1.
2 4
1 ⇒ 
49. Let  = 4 + 1 ⇒  = 4 +
=+
4 
(2 − 1)2
So
2 1)
4 ( −
= 2 ln | + 1| + 12 ln(2 + 1) − tan−1  = (2 ln 2 + 12 ln 2 − 4 ) − (0 + 0 − 0)
0
 1
 


   − 1 
1
11

=
= 5 ln 2 2−+
+  += 2 1  ln
⇒
2
=
2
2
2
2
√
[by Formula 19]

=
=
2
( + 1) ( − 1)
2−−1 1 ( −2 1)   + 1  + 
2 1+ 12 4 ( + 1)
(2 − 1)
( − 1) 
 4 + 1
Section 7.5 Strategy for Integration
4

2
2 √
2
1
1
1 = ( °
+
1)(
− 1)Learning.
+ (
− 1)Reserved.
(
1)(
+or(
+ 1)or2posted
. =
⇒accessible
 =website,
,  in=whole
−1or in⇒
c 2016
 +4
 + 1)
Cengage
All Rights
May
be 1
scanned,
copied,
duplicated,
to a 1
publicly
part.  = 4 .
+−
1not−
4
+
= ln √
4 + 1 + 1 
Equating coefficients of 3 gives  +  = 0, and equating coefficients of 1 gives 1 =  +  −  +  ⇒



1

1 =  + 14 −  + 14 ⇒ √12 =  − . So  = 14and  = − 14 . Therefore,
2  
√
=
50. As in Exercise
49, let  = 4 + 1. Then
. Now
=
8


2
1
2
2
2 4 + 1
( − 1)2
4 ( − 1)  
 

14
14
14
−14

√
+
+
=8
+

1
+ 1+ (
+ 1)2+ − 1+ (
− 1)2 ⇒
2 41+ 1
=
=
2
2
2
2
2
2
( + 1) ( − 1)   + 1
( + 1)
−1
( − 1)
( − 1)

2
2
−2
−2

+
2(
+
1)
+
2(
−
1)
−
=
−
1 + 1)2 .  = 1 ⇒  = 14 ,  = −1 ⇒  = 14 .
1 = ( + 1)( − 1)2 + ( − 1)2 + 1( − 1)( + 1)2 +
(
2
2
= 2 ln
− 2 ln |
− 1| − of 1 gives
+1 = +− + ⇒
Equating coefficients of 3 gives
 |
++
 1|
=−
0, and equating
coefficients
+1
−1
√

− . So  =1 and  =
1 =  + 14 −  + 14 ⇒ 12 = √
2
2 − 14 . Therefore,
= 2 ln 4 + 1 + 1 4− √
− 2 ln 4 + 1 − 1 − √
+
4
+
1
+
1
4
+
1−1

 

14
14
14
−14

√
+
+
=8

√+  1
2 4 + 1 1
2 + 1)2
1

+
1
(
−
1
(
− so
1)2
2

51. Let 2 = tan  ⇒  = 2 tan ,  = 2 sec  , 4 + 1 = sec ,

 
2
2