Thomas Calculus 13th [Solutions]

Section 8.6 Integral Tables and Computer Algebra Systems 599
33.
2
1 sin (cos )
2 2
2 t t dt u sin t 1 u du 2 1 1 u
(cot t) 1 sin t dt ; 1 u ln C
sin t du cost dt u u
(We used FORMULA 31 with a 1 )
2 1 1 sin t
1 sin t ln
sin t
2
C
34.
2
dt
cost dt
; [ u sin t, du cos t dt] du 1 2 4
2 2 2
(tan ) 4 sin (sin ) 4 sin 4
2
ln u
t t t t u u
u
C
(We used FORMULA 34 with a 2)
1 2 4 sin t
ln
2 sin t
2
C
35.
u
ln y
dy u
u
2 2
; y e e du du ln u 3 u C ln ln y 3 (ln y)
C
2 u 2 2
y 3 (ln y) e 3 u 3 u
u
dy e du
(We used FORMULA 20 with a 3 )
36.
t y
2 2 2
1 2 1 1 2 1
tan y dy; y t 2 t tan t dt 2 t tan t 1 t dt t tan t t dt
2 2 1
2 1
2
t
t
dy 2t dt
(We used FORMULA 101 with n 1, a 1)
2
t 2 tan 1 t t 1 dt dt t 2 tan 1 t t tan 1 t C y tan 1 y tan
1 y y C
2 2
t 1 1 t
37.
1 2
dx 1 dx; [ t x 1, dt dx] 1 dt ln t t 4 C
2 2 2
x 2x 5 ( x 1) 4 t 4
(We used FORMULA 20 with a 2 )
2 2
ln ( x 1) ( x 1) 4 C ln ( x 1) x 2x 5 C
38.
2 2 2
2 2
x x t x 2 ( t 2)
dx dx;
dt t 4t 2 dt t dt 4t
dt 4 dt
2 2 2 2 2 2 2
x 4x 5 ( x 2) 1 dt dx t 1 t 1 t 1 t 1 t 1
(We used FORMULA 25 with a 1 ) (We used FORMULA 20 with a 1 )
2
1 2 t t 1 2 2
ln t t 1 4 t 1 4ln t t 1 C
2 2
2
1
2 ( x 2) ( x 2) 1
2 2
ln ( x 2) ( x 2) 1 4 ( x 2) 1 4ln ( x 2) ( x 2) 1 C
2 2
7
2
( x 6) x 4x
5
ln ( x 2) x 4x 5
C
2 2
2
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