微积分英文课件PPT (6)

(2)
xn dx
1 n1
x n1
C
(n 1)
(3)
dx x
ln
x
C
( ln x ) 1
x
(4)
1
dx x
2
arctan
x
C
or arccot x C
(5)
dx arcsin x C 1 x2
or
arccos x C
(6) cos xdx sin x C
(7) sin xdx cos x C
xexdx
Evaluate xexdx
xexdx
exd 1 x2
2
xdex
1 x2 ex 1 x2dex
xex exdx 2
2
xex ex C 1 x2 ex 1 x2exdx
2
2
Example 4 Solution
Evaluate x2exdx
x2exdx x2dex
Example Prove the reduction
sinn xdx 1 cos x sinn1 x n 1 sinn2 xdx
n
n
Where n>2 is integer
Proof
sinn xdx sinn1 xd cos x
sinn1 x cos x cos xd sin n1 x sinn1 x cos x (n 1) cos2 x sin n2 xdx sinn1 x cos x (n 1) (1 sin2 x) sin n2 xdx
x ln 2 x
2[
x
ln
x
x
1dx] x
x ln 2 x 2x ln x 2x C
Example9 ex sin xdx exd cos x
ex cos x ex cos xdx
ex cos x exd sin x
ex cos x ex sin x ex sin xdx
Solution Let u x, dv cos xdx,
Then du dx, v sin x
x cos xdx
x sin x sin xdx
xsin x cos x C
udv uv vdu
Example 2 Evaluate x2 sin xdx
Solution
x2ex ex 2xdx
x2ex 2 xdex
x2ex 2[xex exdx]
x2ex 2xex ex C
ex x2 2x 2 C
Example5
x ln xdx
3
3
ln
x(
2x2 3
)dx
ln
xd
2x2 3
3
3
2x 2
ln x
2x 2 1 dx
33
Thus
sinn xdx 1 cos x sinn1 x n 1 sinn2 xdx
n
n
Chapter 7
Techniques of Integration L’Hospital Rule
and Improper Integrals
7.1 Basic Integration Formulas
Table of Indefinite integrals
(1) dx x C
( C is a constant)
0
0
(ex sin x ) 2 2 ex cos xdx
0
0
(ex sin x ) 2 0
2 cos xdex
0
e2
(ex
cos x ) 2 0
2 ex sin xdx
0
so
2 ex sin xdx
0
e2
1
2 ex sin xdx
0
Thus
2 ex sin xdx
1
e2
1
0
22
3x
2x 2
ln x 33
2 3
xdx
3
2x2 ln x
4x2 C
3
9
Example6 x arctan xdx
arctan
xd
x2 2
arctan x x2 2
wk.baidu.com
x2 2
1
1 x2
dx
x2 2
arctan
x
1 2
1
1
1 x
2
dx
x2 arctan x 1 x arctan x C
sinn1 x cos x (n 1) (1 sin2 x) sin n2 xdx sinn1 x cos x (n 1) sin n2 xdx
(n 1) sinn xdx
Therefore
n sinn xdx cos x sinn1 x (n 1) sinn2 xdx
( f (x)g(x))dx f (x)g(x)dx f (x)g(x)dx
f (x)g(x)dx ( f (x)g(x))dx f (x)g(x)dx f (x)g(x)dx f (x)g(x) f (x)g(x)dx
The formula for integration by parts
so
ex sin xdx
1 ex sin x cos x C
2
or
ex sin xdx
sin xdex
ex sin x ex cos xdx
ex sin x cos xdex
ex sin x ex cos x ex sin xdx so, ex sin xdx 1 ex sin x cos x C
2
2
1 x2 1arctan x 1 x C
2
2
Example 7 arccos xdx
x arccos x
x dx
1 x2
x arccos x 1 x2 C
Example 8 ln 2 xdx
xd ln2 x
x ln 2 x
x
2
ln
x
1 x
dx
x ln 2 x 2 ln xdx
2
Integration by parts can also be usd in connection With definite integrals,the formula is
b
f (x)g(x)dx
a
f (x)g(x)]ba
b
f (x)g(x)dx
a
b
b
a [ f (x)g(x)] dx a ( f (x)g(x) f (x)g(x))dx
f (x)g(x)]ba
b
f (x)g(x)dx
a
b
f (x)g(x)dx
a
b f (x)g(x)dx
a
f (x)g(x)]ba
b
f (x)g(x)dx
a
or
b udv
a
uv]ba
b
vdu
a
1
1
Exampe
arctan xdx. arctan xdx
0
0
x arctanx 1 0
ln a
(14) sh xdx ch x C
sh x ex ex 2
ch x ex ex 2
(15) ch xdx sh x C
Exercise:
2x 9
1.
dx x2 9x 1
2.
dx 8x x2
3. (secx tan x)2 dx
Exercise:
/4
4. 0 1 4 cos4xdx
1x 0 1 x2 dx
1 ln(1 x2 ) 1 1 ln 2.
42
0 42
e
Example x ln xdx.
1 e ln xdx2
1
21
1 x2 ln x 2
e 1 e x2 1 dx 1 21 x
e2 2
e2 1 4
1 x2 e
4
1
Example 2 ex sin xdx 2 sin xdex
udv uv vdu
x2 sin xdx x2d cos x
x2 cos x ccoossxxd(x22x)dx
x2 cos x 2 xd sin x
x2 cos x 2(x sin x sin xdx)
x2 cos x 2xsin x 2cos x C
Example 3 Solution
(8)
dx cos 2
x
sec2
xdx
tan x C
(9)
d sin
x
2
x
csc2
xdx
cot
xC
(10) sec x tan xdx sec x C
(11) csc x cot xdx csc x C
(12) ex dx ex C (13) a xdx a x C
3x 2
5.
dx 1 x2
6.
3x2 3x
7x 2
dx
7. sec xdx
7.2 Integration by Parts
Integration by Parts If f (x) and g(x) are differentiable functions ,then
( f (x)g(x)) f (x)g(x) f (x)g(x)
f (x)g(x)dx f (x)g(x) f (x)g(x)dx
Let u f (x) and v g(x)
Then the formula for integration by parts becomes
udv uv vdu
Example 1 Find x cos xdx