Page 206 - Calculus Workbook For Dummies

190 Part IV: Integration and Infinite Series

π
q # π sec 5 - πh tan 5 - πh dt = 2
5
t
t
^
^
π
0
Don’t let all those 5s and s distract you — they’re just a smoke screen.
π
π $
Guess: sec 5 - πh. Diff: sec 5 - πh tan 5 - h 5.
t
t
t
^
^
^
1 1
π $
t
t
t
Tweak: π sec 5 - πh. Diff: π sec 5 - πh tan 5 - h 5. Bingo. So now —
^
^
^
π
1 1 2
t
^
h
^
π sec 5 - πhE = π 8 sec 4 ^ π - sec - πhB = π
0
r # . 4 5 dx = 3 tan 3 + C
1
-
x
1 + 9 x 2 2
I bet you’ve got the method down by now: Guess, diff, tweak, diff; Guess, diff, tweak, diff. . . .
1
1
-
x
Guess: tan 3 . Diff: 2 3 $ .
1 + ^ 3 xh
3 3 1
1
-
Tweak: tan 3 . Diff: $ 3 $ . That’s it.
x
2 2 1 + ^ 3 xh 2
s # sinx dx = - 2 cos x + C
cosx
1. It’s not plain old x , so substitute u = cosx.
2. Differentiate and solve for du.
du
dx = - sinx
du = - sinxdx
3. Tweak inside and outside of integral: - # - sinx dx
4. Pull the switch: = - # du cosx
u
5. Antidifferentiate with reverse power rule: = - # u - / 1 2 du = - 2 u + C
/ 1 2
/ 1 2
-
6. Get rid of u: 2^ cosxh + C = - 2 cosx + C
5
5
t # x 4 3 2 x + 6 dx = 3 _ x + 3i 3 2 x + 6 + C
5
20
5
1. It’s not plain old x , so substitute u 2 x + 6.
=
2. Differentiate and solve for du.
du 4
dx = 10 x
4
=
du 10 x dx
1 # 5
3. Tweak inside and outside: 10 x 4 3 2 x + 6 dx
10
1 #
4. Flip the switch: = 3 u du
10
3
1 3 / 4 3 3 u u
5. Apply the power rule in reverse: = $ u + C = + C
10 4 40
5
5
5
5
3 2 x + 6i 3 2 x + 6 3 _ x + 3i 3 2 x + 6
_
6. Switch back: + C = + C
40 20