2arcsin( ln x )
e
. 6
1 Example: Find dx . (a 0) 2 2 0 x a x Solution: Let x a sin t , dx a cos tdt , x a t , x 0 t 0, 2
x
7
Example: Find
3 e4
e
3 e4
dx . x ln x(1 ln x )
Solution

e
d (ln x ) ln x(1 ln x ) d (ln x ) 2 ln x (1 ln x )
3 e4
3 e4
3 e4
e
e
d ln x 1 ( ln x )2

2
when x 1 ; thus by the formula, we have

1
0
1 x 2 dx 2 cos 2 tdt
0

1 1 2 t sin 2t 2 2 0


.
4
5
Integration by Substitutions for definite integrals
17
Integration by Parts
Integration by parts for definite integrals
When u and v are differentiable functions of x on the interval [a , b]. Then

Example: Evaluate
2 2
f (cos t )dt f (cos x )dx;
0 0
例 若 f ( x ) 在[0,1]上连续，证明
0

xf (sin x )dx f (sin x )dx . 2 0
x t dx dt ,
Proof: Let

x 0 t ,
School of Science, BUPT
Integration by substitution and by parts in definite integrals
2
Integration by Substitutions for definite integrals
When we want to evaluate the value of a definite integral can find the corresponding indefinite integral
d dv du ( uv ) u v dx dx dx
Integrating both sides with respect to x and rearranging leads to the integral equation
dv d du u dx ( uv ) dx v dx dx dx dx du uv v dx . dx
2
0
2 cos n tdt
0

2 cos n xdx .
0

6
Integration by Substitutions for definite integrals
1 Example: Find lim 2 x 0 x sin xt 0 t dt du u . t 0 u 0; t x u x 2 . Solution Let xt u, then t , dt x x Then x 2 sin u du x sin xt x 2 sin u 0 t dt 0 u x 0 u du x So 0 2 x sin u 0 sin x 2 0 u du x 2 2 x 1 x sin xt lim 2 dt lim 1. lim 2 0 x 0 x x 0 x 0 t x 2x
偶倍奇零
(1) If (2) If Proof
a 0
then

a
a
f ( x )dx 2 f ( x )d x
0
a
then

a
a
a
f ( x )d x 0
a
a f ( x) dx a f ( x) dx 0 f ( x) dx
f (t ) d t f ( x) dx [ f ( x ) f ( x ) ] dx
b
a
udv uv a vdu
b a
b

4
0
e x dx .
Solution Let
2 x t , then x t , dx 2tdt . Thus
4

0
e dx e 2tdt 2 tde 2 te
x t t 0 0
2
2

t 2 0
e t dt
0
2 0


2

0
f (sin x )dx
Integration by Parts
Integration by parts for indefinite integrals
When u and v are differentiable functions of x, the Product Rule for differentiation tell us that

b
a
f ( x )dx , if we
f ( x )dx , then we can use
the Newton-Leibniz formula to obtain the value immediately.
Theorem (Integration by substitution for definite integrals) Suppose that the function x ( t ) is continuously differentiable on the interval [ , ] , ( ) a , ( ) b , and that f is continuous in the range of
0 0 a 0 a
Let x t

f ( x ) f ( x ) f ( x ) f ( x )
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Example: Find Solution:
1
1
2 x 2 x cos x dx . 2 1 1 x
1
1
1
1 x cos x 2x2 dx dx 1 2 2 1 1 x 1 1 x
, v x,

1 2
0
arcsin xdx x arcsin x 0
1 2
1 2
xdx 1 x2
0
1 1 1 1 2 d (1 x 2 ) 2 6 2 0 1 x2
, I . Then

b
a
f ( x )dx f ( t ) ( t )dt .


3
Integration by Substitutions for definite integrals
Proof Let F be an antiderivative of f on the interval I, then
例 若 f ( x ) 在[0,1]上连续，证明
0
2
f (sin x )dx f (cos x )dx .
0
2
Proof （1）设 x t 2 x 0 t , 2
dx dt,
0
2
x t 0, 2 0 f (sin x )dx f sin t dt 2 2
b
Since

a
f ( x )dx F (b) F (a )
d F ( t ) f ( t ) ( t ), dt
we have


f ( t ) ( t )dt F ( t )

F ( ) F ( )
F (b) F (a )
1 d (cos x ) arctan(cos x )0 2 2 0 1 cos x 2 2 ( ) . 4 2 4 4
xf (sin x )dx f (sin x )dx . 2 0


0
(1) If f ( x )=f ( x )， then
0
T


a
0
f ( x )dx f (a x )dx
0
a
(let t a x )

2 0
f (sin x )dx 2 f (cos x )dx
0


0
f (sin x )dx 2 2 f (sin x )dx
0

(7)


0
xf (sin x )dx f (sin x )dx



0
0
xf (sin x )dx f (sin x )dx. 2 0
0
Example: Find
Solution
0

x sin x dx 2 1 cos x
0

x sin x sin x dx dx 2 2 2 0 1 cos x 1 cos x