Notes on Integration involving the Natural Logarithm Function

# Notes on Integration involving the Natural Logarithm Function

This preview shows pages 1–4. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Notes on Integration involving the Natural Logarithm Function As you learned in calculus I, differentiation formulas read backwards yield integration formulas, recall for example sin cos cos sin d x x xdx x C dx = = + . Similarly 1 ln ln d dx x x C dx x x = = + where C is the constant of integration. This fact opens up a wide variety of problems that were not possible before. The examples below illustrate some of the possibilities. Example 1 : Evaluate the indefinite integral 2 3 1 t dt t + . Solution : 2 2 3 3 1 1 ln 2 t t dt t dt t C t t-- + = + = + + - 2 1 ln 2 t C t =- + Example 2 : Evaluate the indefinite integral ( 29 2 ln dx x x + . Solution : ( 29 1 2 ln 2 ln dx dx x x x x = + + 2 ln dx u x du x = + = ln ln 2 ln du u C x C u = = + = + + ( The next example is a definite integral. When changing the variable of integration, dont forget to change the limits of integration also. ) Example 3 : Evaluate 4 cos 2 3 sin 2 d - . Solution : ( 29 4 4 cos 2 1 cos 2 3 sin 2 3 sin 2 d d = -- { 3 sin 2 2cos2 u du d =- = - 2 3 3 2 3 2 1 1 1 1 ln 2 2 2 du du u u u =- = = ( 29 1 1 3 ln3 ln 2 ln 2 2 2 =- = 1 Example 4 : Evaluate 3 4 1 2 1 dx x x + - + . Solution : 3 4 3 4 1 2 1 1 2 1 dx dx dx x x x x + = + - +- + { 1 u x du dx = - = - { 2 1 2 v x dv dx = + = ( 29 1 1 3 4 2 3 2 du dv du dv u v v u- = + =- 2ln 3ln 2ln 2 1 3ln 1 v u C x x C =- + = +-- + 2 3 ln 2 1 ln 1 x x C = +-- + 2 3 2 1 ln 1 x C x + = + - ( 29 2 3 2 1 ln 1 x C x + = +- . In Example 4 the absolute value sign was dropped for the term ( 29 2 2 1 x + which justified by the fact that this term is always positive. An interesting idea is to note that C is a constant. So there is a numbers k for which ln C k = . Therefore we could write the answer as ( 29 ( 29 2 2 3 3 2 1 2 1 ln ln ln 1 1 x x C k x x + + + = +-- ( 29 2 3 2 1 ln 1 k x x + = - . Example 5 : Evaluate tan xdx Solution : sin tan cos x xdx dx x = { cos sin u x du xdx = = - du du u u- = = - ( 29 1 ln ln cos ln cos u C x C x C- = - + = - + = + 1 ln ln sec cos C x C x = + = + . Example 6 : Evaluate sec xdx . Solution : ( 29 sec sec tan sec tan sec x x x xdx dx x x + = + ln ln tan sec du u C x x C u = = + = + + . ( 29 { 2 tan sec sec sec tan u x x du x x x dx = + = + The results of Examples 5 and 6 will come up often enough in the remaining part of this course that you should commit them to memory. In your homework assignment, you will derive corresponding results for cot xdx and csc xdx . Add these to your 2 memory also. Congratulations!! You now know how to integrate all the trigonometric functions!...
View Full Document

## This note was uploaded on 05/06/2011 for the course MATH 256 taught by Professor Buekman during the Spring '11 term at Purdue University-West Lafayette.

### Page1 / 8

Notes on Integration involving the Natural Logarithm Function

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online