5.2 The Natural Logarithmic Function: Integration

After this lesson, you should be able to:

Use the Log Rule for Integration to integrate a rational function.Integrate trigonometric functions.

Log Rule for Integration

Let u be a differentiable function of x.

Because du = u ’dx the 2nd formula can be written as

cuduu

cxdxx

ln1

.2ln1

.1

cudxu

uln

'

In the last section, we have already learned that

xx

dx

d 1ln 0x

)(

)()(ln

xf

xfxf

dx

d 0)( xf1. 2.

It is easily concluded:

Examples for Log Rule

let u = 4x – 1, du = 4dx and x = (u +1)/4, then

cxdxx

dxx

ln21

22

.1

dx

x 14

1.2

duu

dxx

1

4

14

14

1

4

1cxcu 14ln

4

1ln

4

1

Examples of Quotient Form of the Log Rule

13,,ln13

.3 2333

2

xduxxuLetcxxdxxx

x

dxxduxunotecxdxx

x 22

sec,tan:tanlntan

sec.4

dxxduxxunote

cxxdxxx

xdxxx

xdxxx

x

22,2:

2ln2

1

2

22

2

1

2

12

2

1

2

1.5

2

2222

dxduxunote

xdxx

dxx

3,23:

23ln3

1

23

3

3

1

23

1.6

1( )

5 2f x

x

Example 7 Find the area bounded by the following function, the x axis and the lines x = –1 and x =1:

0.424 sq units

Solution

)25(25

1

2

1

25

2

2

1

25

1 1

1

1

1

1

1

xdx

dxx

dxx

A

]1 ,1[ ,025

1)(

x

xxf

Note that

3

7ln

2

1)25ln(

2

11

1

x

Example 8 Find the antiderivative of

Solution

dxx

xxdxx

xx

2

1062

2

22 23

Note that

dxx

xx

2

223

2

1062

2

22 23

xxx

x

xx

dx

xdxxx

2

10622

)2(2

1106

3

1 23 xdx

xxx

Cxxxx |2|ln1063

1 23

Example 9 Find the antiderivative of

Solution

du

uudu

u

udu

u

udx

x

x2222

112

)1(2

)1(2

)1(

2

Let u = x + 1, then du = dx and x = u – 1

dxx

x 2)1(

2

duu

duu 2

12

12 duudu

u22

12

Cu

u

12||ln2

1

Cu

u 2

||ln2

Cx

x

1

2|1|ln2

Guidelines for Integration

Summary:• Memorize the form• Make the problem fit the form• Integrate

Example 10 Find the antiderivative of

Solution

)(lnln

1

ln

1xd

xdxxx

We notice that 1/x dx = d(lnx)

dxxx ln

1

Apply the Log Rule for u = lnx

Cx |ln|ln

Integrals of the Six Trig Functions

cuuudu

cuuudu

cuuducuudu

cuuducuduu

cotcsclncsc.6

tanseclnsec.5

sinlncot.4coslntan.3

sincos.2cossin.1

Example 11 Find the antiderivative of

Solution

dxxx

xxxdxxx

xxx

tansec

tansecsec

tansec

tansecsec

2

We multiply and divide a factor (sec x + tan x) on the integrand, and we also notice that

d(tan x) = sec2x dx and d(sec x) = sec x tan x dx

dxxsec

dxxxxxxd )tansec(sec)tan(sec 2

Cxxxxdxx

|tansec|ln)tan(sectansec

1

Example 11 Find the average value of

on the interval of [0, π/4]

Solution

4

0

4

0

24

0

2 sec4

sec4

tan14

Value Average

xdxdxxdxx

xxf 2tan1)(

Notice that sec x > 0 on the interval [0, π/4]

)12ln(4

|tansec|ln4 4/

0

xx

Homework

Pg. 338 1-15 odd, 27-39 odd, 43-51 odd, 71