Thomas Calculus 13th [Solutions]

Chapter 8 Additional and Advanced Exercises 651
3.
sin 1
, dx ; , x
2
1
2
;
x
u x du dv x dx v
x
dx
sin
cos d
2
2 2
sin 1 x 1
2
sin x dx ;
2
2 1 x
x x dx x
2 2
2
1 1 sin cos
1 1 2
sin x
d
sin x sin sin
2 2cos 2 2
x x dx x x d
2 2 2
2 1
x 1 1 sin 2 1 sin cos
1 1 sin
sin x sin x
x x x
sin
2 2 2 4 2 4 2 4
x C x C x C
4.
1
sin ; y dy
z
dz
y
dy
2 y
1
2 sin ;
z
z dz
from Exercise 3,
2 1
2 1
1 sin z 1 z sin z
z sin z dz
z z
C
2 4
1 2 1
1 1 y 1 y sin y 1 y y sin y
sin sin sin
2 2 2
y dy y y C y y C
5.
u tan
dt
t sin
cos d
;
d
2
; du sec d
du
2
t 1 t dt cos d sin cos tan 1
2
( u 1) u 1
d
du
2
u 1
1 du 1 du 1 u du 1 1 1 1
ln u tan 1 ln tan 1 1
u u 1 u 1 u 1
u C C
2 1 2 2 2 2 2 2 2 2 sec 2
1 2 1 1
ln 1 sin
2 2
t t t C
1 1 1 1 2x
2 2 2x
2 2
x 4 x 2 4x
x 2x 2 x 2x
2
x 2x 2 ( x 1) 1 x 2x 2 ( x 1) 1
dx dx dx dx
6.
4
2
2
2
2 2
16 2 2 2 2
2
1 2 2 1 1 1
ln x x tan ( 1) tan ( 1)
x 2x
2
16 2 8
x x C
x
x
7. lim sin t dt lim cost lim cos x cos( x) lim cos x cos x lim 0 0
x x x
x
x x x
8. lim 1 cos t
dt ; 2
x 0 x t
1
t
2
lim lim 1
cost
1 lim
t 0 t 0 x 0
cost
t
2
1 cos t
1
dt diverges since dt
x
2
2
t
0 t
diverges; thus
1
lim x cos t dt is an indeterminate 0 form and we apply l Hôpital s rule:
2
x 0 x t
x cost
t
1
t
2
2 1
x
cos x
x
2
1
lim x dt
cos lim lim lim cos 1
0 x dt x
x t x 0 x 0 x 0
1
x
2
9.
n
n
lim ln 1 k lim ln 1 1 1
1
0
ln 1 ; u 1 x,
du dx
n
k
x dx
n
n n n
k 1
n
x 0 u 1, x 1 u 2
k 1
2 2
ln
1 u du u ln u u 2ln 2 2 ln1 1 2ln 2 1 ln 4 1
1
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