Sample Problems. Practice Problems
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1 Lecture Notes Partial Fractions page Sample Problems Compute each of the following integrals.. x dx. x + x (x + ) (x ) (x ) dx 8. x x dx... x (x + ) (x + ) dx x + x x dx x + x x + 6x x dx + x x (x ) dx x + (x ) dx sec x dx csch x dx Practice Problems. x + x dx 7. x x 0x x dx. x + x + dx.... x x x 8 dx x a dx x x dx x x + dx x 7 x 6x + 9 dx x + 7x + x + dx x x + 0 (x ) (x + 9) dx... x + x (x + 6) dx x + 6 (x + ) dx x x + 9 dx 6. x x + x dx. x x 6 dx 6. x + x x x dx c copyright Hidegkuti, Powell, 0 Last revised: February, 0
2 Lecture Notes Partial Fractions page Sample Problems - nswers.) ln jx j ln jx + j + C.) ln jx + j ln jx + j + C.) ln jx j ln jx + j + C.) x 6.) ln jx j x + x + ln jx + j + ln jx j + C.) 6 ln jx + j ln jx j + ln jx j + C 9 x + C 7.) x (x ) + C x (x ) + C 8.) x arctan x ln jx + j + ln jx j + C 9.) ln jsec x + tan xj + C ln jsec x tan xj + C 0.) ln je x j ln (e x + ) + C Practice Problems - nswers.) ln jxj ln jx + j + C.) 7 6 ln jx + j 6 ln jx j + C.) a ln jx aj ln jx + aj + C a.) ln jx j + ln jx + j + C.) ln x + tan x + C 6.) x + ln jx j 9 ln jx + j + C 7.) x x ln jx j + ln jx + j + C 8.) ln jx j + x + C 9.) x + 7 ln x + + tan x + C 0.) ln jx j tan x + C.) x + ln jx j ln jx + j tan x + C.) tan x + ln x + + C.) ln x + 6x + C.) ln jx + j 9 x + + C.) ln x + 9 tan x + C 6.) ln jxj + ln jx j ln jx + j + C c copyright Hidegkuti, Powell, 0 Last revised: February, 0
3 Lecture Notes Partial Fractions page Sample Problems - Solutions Compute each of the following integrals.. x dx Solution: We factor the denominator: x (x + ) (x ). Next, we re-write the fraction x as a sum (or di erence) of fractions with denominators x + and x. This means that we need to solve for and B in the equation x + + B x x To simplify the left-hand side, we bring the fractions to the common denominator: (x ) B (x + ) x + Bx + B + (x + ) (x ) (x ) (x + ) x ( + B) x + B x Thus we have We clear the denominators by multiplication ( + B) x + B x x ( + B) x + B The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. In other words, ( + B) x + B 0x + This gives us an equation for each coe cient, forming a system of linear equations: We solve this system and obtain and B. + B 0 + B So our fraction, x x + + x Now we can easily integrate: can be re-written as x + +. We check: x (x ) (x + ) (x + ) (x ) + (x ) (x + ) x (x ) + (x + ) (x + ) (x ) x + + x + (x + ) (x ) x dx x + + x dx x + dx + x dx ln jx + j + ln jx j + C c copyright Hidegkuti, Powell, 0 Last revised: February, 0
4 Lecture Notes Partial Fractions page Method : The values of and B can be found using a slightly di erent method as follows. Consider rst the equation x + + B x x We bring the fractions to the common denominator: and then multiply both sides by the denominator: (x ) B (x + ) + (x + ) (x ) (x ) (x + ) x (x ) + B (x + ) The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute (0) + B () B Let us substitute x into both sides: ( ) + B (0). and so and B. x (x + ) (x + ) dx x Solution: We re-write the fraction as a sum (or di erence) of fractions with denominators (x + ) (x + ) x + and x +. This means that we need to solve for and B in the equation x + + B x + x (x + ) (x + ) To simplify the left-hand side, we bring the fractions to the common denominator: x + + B x + (x + ) (x + ) (x + ) + B (x + ) (x + ) (x + ) ( + B) x + + B (x + ) (x + ) (x + ) + B (x + ) (x + ) (x + ) x + + Bx + B (x + ) (x + ) Thus we have We clear the denominators by multiplication ( + B) x + + B (x + ) (x + ) x (x + ) (x + ) ( + B) x + + B x The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. In other words, ( + B) x + + B x + 0 c copyright Hidegkuti, Powell, 0 Last revised: February, 0
5 Lecture Notes Partial Fractions page This gives us an equation for each coe cient, forming a system of linear equations: We solve this system and obtain and B. So our fraction, x + + x + x (x + ) (x + ) Now we can easily integrate: can be re-written as + B + B 0 x + + x +. We check: (x + ) (x + ) (x + ) (x + ) + (x + ) (x + ) (x + ) 9 (x + ) x + x (x + ) (x + ) (x + ) (x + ) x (x + ) (x + ) x (x + ) (x + ) dx x + + x + dx x + dx The second integral can be computed using the substitution u x +. x + dx ln jx + j ln jx + j + C Method : The values of and B can be found using a slightly di erent method as follows. Consider rst the equation x + + B x + x (x + ) (x + ) We bring the fractions to the common denominator: and then multiply both sides by the denominator: (x + ) (x + ) (x + ) + B (x + ) (x + ) (x + ) x (x + ) (x + ) (x + ) + B (x + ) x The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: (0) + B + 8 B B Let us substitute x into both sides: Let us substitute ( ( ) + ) + B ( + ) ( ) 8 + B (0) and so and B. c copyright Hidegkuti, Powell, 0 Last revised: February, 0
6 Lecture Notes Partial Fractions page 6. x + x x dx Solution: We factor the denominator: x x + x (x + ) (x ). Next, we re-write the fraction x x as a sum (or di erence) of fractions with denominators x + and x. This means that we need to solve for and B in the equation x + + B x x + x x To simplify the left-hand side, we bring the fractions to the common denominator: (x ) B (x + ) x + Bx + B + (x + ) (x ) (x ) (x + ) x x Thus ( + B) x + B x x We clear the denominators by multiplication x + x x ( + B) x + B x x ( + B) x + B x + The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. This gives us an equation for each coe cient, forming a system of linear equations: + B + B We solve the system and obtain and B. x + So we have that our fraction, x x can be re-written as x + +. We check: x x + + x (x ) (x + ) (x ) + (x + ) (x ) + (x + ) x + + x + (x ) (x + ) (x + ) (x ) x x Now we can easily integrate: x + x x dx x + + x dx x + dx + x x + x x dx ln jx + j + ln jx j + C Method : The values of and B can be found using a slightly di erent method as follows. Consider rst the equation x + + B x x + x x We bring the fractions to the common denominator: and then multiply both sides by the denominator: (x ) B (x + ) + (x ) (x + ) (x ) (x + ) x + x x (x ) + B (x + ) x + The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute (0) + B () + B 8 B c copyright Hidegkuti, Powell, 0 Last revised: February, 0
7 Lecture Notes Partial Fractions page 7 Let us substitute x into both sides: ( ) + B (0) + and so and B.. x + x x + 6x x + x dx Solution: This rational function is an improper fraction since the numerator has a higher degree than the denominator. We rst perform long division. This process is similar to long division among numbers. For example, to simplify 8 ; we perform the long division 8 7 R which is the same thing as to say that The division: x x + x + x ) x + x x + 6x x x + x x x + 6x x + 6x 8x x + 8x x x + 0 x x Step : x x x x + x x + x x x + x x x x + x We add that to the original polynomial shown above. x Step : x x x x + x x 6x + 8x x 6x + 8x x + 6x 8x We add that to the original polynomial shown above. Step : x x x + x x + x 0 x + x 0 x x + 0 We add that to the original polynomial shown above. The result of this computation is that very much like Thus x + x x + 6x x dx + x x + x x + 6x x + x x We apply the method of partial fractions to compute x x + + x x + x x x + + x x + x dx x x + x + C + x + x dx x x + x dx. x x + dx + x x + x dx c copyright Hidegkuti, Powell, 0 Last revised: February, 0
8 Lecture Notes Partial Fractions page 8 We factor the denominator: x x + x (x + ) (x ). Next, we re-write the fraction x + x as a sum (or di erence) of fractions with denominators x + and x. This means that we need to solve for and B in the equation x + + B x x x + x To simplify the left-hand side, we bring the fractions to the common denominator: (x ) B (x + ) x + Bx + B + (x + ) (x ) (x + ) (x ) x + x ( + B) x + B x + x Thus ( + B) x + B x + x We clear the denominators by multiplication x x + x ( + B) x + B x The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. This gives us an equation for each coe cient that forms a system of linear equations: We solve the system and obtain and B. So our fraction, x x + x x + + x Now we can easily integrate: x x + x dx can be re-written as Thus the nal answer is x + x x + 6x x + x + B + B x + +. We check: x (x ) (x + ) (x + ) (x ) + (x ) (x + ) (x ) + (x + ) (x + ) (x ) x + x + 8 x (x + ) (x ) x + x x x + x x + + dx x dx x + dx + x x + dx + x dx dx ln jx + j + ln jx j + C x x x + x + C + x x + x dx x x + x + ln jx + j + ln jx j + C x + x + C + ln jx + j + ln jx j + C c copyright Hidegkuti, Powell, 0 Last revised: February, 0
9 Lecture Notes Partial Fractions page 9 Method : The values of and B can be found using a slightly di erent method as follows. Consider rst the equation x + + B x x x + x We bring the fractions to the common denominator: and then multiply both sides by the denominator: (x ) B (x + ) + (x + ) (x ) (x + ) (x ) x (x + ) (x ) (x ) + B (x + ) x The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute ( ) + B ( + ) x 0 + B B B Let us substitute x into both sides: ( ) + B ( + ) ( ). and so and B. x + x (x + ) (x ) (x ) dx x + x Solution: We re-write the fraction as a sum (or di erence) of fractions with denominators x +, x and x. This means that we need to solve for, B, and C in the (x + ) (x ) (x ) equation x + + B x + C x x + x (x + ) (x ) (x ) To simplify the left-hand side, we bring the fractions to the common denominator: x + + B x + C x (x ) (x ) B (x + ) (x ) C (x + ) (x ) + + (x + ) (x ) (x ) (x + ) (x ) (x ) (x + ) (x ) (x ) x 7x + 0 (x + ) (x ) (x ) + B x x (x + ) (x ) (x ) + C x x (x + ) (x ) (x ) x 7x B x x + C x x (x + ) (x ) (x ) x 7x Bx Bx B + Cx Cx C (x + ) (x ) (x ) ( + B + C) x + ( 7 B C) x + 0 B C (x + ) (x ) (x ) c copyright Hidegkuti, Powell, 0 Last revised: February, 0
10 Lecture Notes Partial Fractions page 0 Thus ( + B + C) x + ( 7 B C) x + 0 B C (x + ) (x ) (x ) We clear the denominators by multiplication x + x (x + ) (x ) (x ) ( + B + C) x + ( 7 B C) x + 0 B C x + x The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. We have an equation for each coe cient that gives us a system of linear equations: + B + C 7 B C 0 B C We solve the system by elimination: rst we will eliminate C from the second and third equations. eliminate C from the second equation, we simply add the rst and second equations. + B + C 7 B C 6 B To eliminate C from the third equation, we multiply the rst equation by and add that to the third equation. + B + C 0 B C B We now have a system of linear equations in two variables: 6 B B We will eliminate B by adding the opposite of the rst equation to the second equation. 6 + B B 8 6 Using the equation 6 + B we can now solve for B. 6 + B 6 + B Using the rst equation, we can now solve for C. 6 + B B + B + C + C + C C To c copyright Hidegkuti, Powell, 0 Last revised: February, 0
11 Lecture Notes Partial Fractions page Thus 6, B, and C So we have that our fraction, 6 x + + x + x Now we can easily integrate: x + x (x + ) (x ) (x ) can be re-written as 6 x + + x +. We check: x (x ) (x ) (x + ) (x ) (x + ) (x ) 6 (x + ) (x ) (x ) + (x + ) (x ) (x ) + (x + ) (x ) (x ) 6 (x ) (x ) (x + ) (x ) + (x + ) (x ) (x + ) (x ) (x ) 6 x 7x + 0 x x + x x (x + ) (x ) (x ) 6 x x x + x + + x x (x + ) (x ) (x ) x x + (x + ) (x ) (x ) x + x (x + ) (x ) (x ) x + x (x + ) (x ) (x ) dx 6 x x + x + dx x dx x dx + x dx 6 ln jx + j ln jx j + ln jx j + C Method : The values of ; B, and C can be found using a slightly di erent method as follows. Consider rst the equation x + + B x + C x x + x (x + ) (x ) (x ) We bring the fractions to the common denominator: (x ) (x ) B (x + ) (x ) C (x + ) (x ) + + (x + ) (x ) (x ) (x + ) (x ) (x ) (x + ) (x ) (x ) x + x (x + ) (x ) (x ) and then multiply both sides by the denominator: (x ) (x ) + B (x + ) (x ) + C (x + ) (x ) x + x The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute ( ) ( ) + B ( + ) ( ) + C ( + ) ( ) + 0 9B + 0C 9B B c copyright Hidegkuti, Powell, 0 Last revised: February, 0
12 Lecture Notes Partial Fractions page Let us substitute x into both sides: ( ) ( ) + B ( + ) ( ) + C ( + ) ( ) ( ) + ( ) Let us substitute x into both sides: ( ) ( 6) + 0B + 0C 8 6 ( ) ( ) + B ( + ) ( ) + C ( + ) ( ) + (0) + B (0) + C (6) () 7 8C 7 C and so 6, B, and C. 6. x (x ) dx Solution: We will re-write the fraction x as a sum (or di erence) of fractions with denominators x (x ) and (x ). This means that we need to solve for and B in the equation x + B (x ) x (x ) To simplify the left-hand side, we bring the fractions to the common denominator: Thus we have x + B (x ) (x ) (x ) + B (x ) + B x + B (x ) (x ) (x ) We clear the denominators by multiplication x + B (x ) x (x ) x + B x The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. This gives us an equation for each coe cient, forming a system of linear equations: + B We solve this system and obtain and B 9. x So our fraction, (x ) can be re-written as x + 9. We check: (x ) x + 9 (x ) (x ) (x ) + 9 x (x ) (x ) x (x ) c copyright Hidegkuti, Powell, 0 Last revised: February, 0
13 Lecture Notes Partial Fractions page Now we can easily integrate: x (x ) dx x + 9 (x ) dx x dx + 9 (x ) dx ln jx j 9 x + C Method : The values of and B can be found using a slightly di erent method as follows. Consider rst the equation x + B (x ) x (x ) We bring the fractions to the common denominator: and then multiply both sides by the denominator: (x ) (x ) + B (x ) x (x ) (x ) + B x The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute (0) + B 9 B 9 The other value of x can be arbitrarily chosen. (There is no value that would eliminate B from the equation.) For easy substitution, let us substitute x 0 into both sides and also substitute B 9: ( ) and so and B 9. x + (x ) dx Solution: We re-write the fraction x + as a sum (or di erence) of fractions with denominators x, (x ) (x ) and (x ). This means that we need to solve for, B, and C in the equation x + B (x ) + C (x ) x + (x ) To simplify the left-hand side, we bring the fractions to the common denominator: x + B (x ) + C (x ) (x ) B (x ) (x ) + (x ) + C (x ) (x ) + B (x ) + C (x ) x x + + B (x ) + C (x ) x x + + Bx B + C (x ) x + ( + B) x + B + C (x ) c copyright Hidegkuti, Powell, 0 Last revised: February, 0
14 Lecture Notes Partial Fractions page Thus We clear the denominators by multiplication x + ( + B) x + B + C (x ) x + (x ) x + ( + B) x + B + C x + The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. We have an equation for each coe cient that gives us a system of linear equations: 0 + B B + C Since 0; this is really a system in two variables: B B + C We solve this system and obtain B and C. x + So our fraction, (x ) can be re-written as (x ) +. We check: (x ) Now we can easily integrate: x + (x ) dx (x ) + (x ) (x ) (x ) + (x ) x + (x ) x + (x ) Both nal answers are acceptable. (x ) + (x ) dx x (x ) + C x (x ) dx + (x ) + C (x ) (x ) (x ) + C x + (x ) + C (x ) dx x (x ) + C Method : The values of ; B, and C can be found using a slightly di erent method as follows. Consider rst the equation x + B (x ) + C (x ) x + (x ) We bring the fractions to the common denominator: and then multiply both sides by the denominator: (x ) B (x ) (x ) + (x ) + C (x ) x + (x ) (x ) + B (x ) + C x + The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute (0) + B (0) + C + C c copyright Hidegkuti, Powell, 0 Last revised: February, 0
15 Lecture Notes Partial Fractions page There is no value other than that would eliminate or B from the equation. Our method will still work. For easy substitution, let us substitute x 0 into both sides and also substitute C : Let us substitute x into both sides: We now solve the system of equations (x ) + B (x ) + C x + (0 ) + B (0 ) B + B (x ) + B (x ) + C x + ( ) + B ( ) B + B + B + B and obtain 0 and B. Recall that we already have C. x 8. x dx Solution: This rational function is an improper fraction since the numerator has the same degree as the denominator. We rst perform long division. This one is an easy one; the method featured below is called smuggling. x x x + x x x + x + x Thus x x dx + x dx dx + x dx x + C + dx We apply the method of partial fractions to compute x dx. We factor the denominator: x x + (x + ) (x ). Next, we re-write the fraction x as a sum (or di erence) of fractions with denominators x + and x +, and x. In the fraction with quadratic denominator, the numerator is linear. This means that we need to solve for and B in the equation x + B x + + C x + + D x x To simplify the left-hand side, we bring the fractions to the common denominator: x + B x + + C x + + D (x + B) (x + ) (x ) x (x + ) (x + ) (x ) + C x + (x ) (x + ) (x + ) (x ) + D x + (x + ) (x + ) (x + ) (x ) x (x + B) (x + ) (x ) + C x + (x ) + D x + (x + ) x (x + B) x + C x x + x + D x + x + x + x x + Bx x B + Cx Cx + Cx C + Dx + Dx + Dx + D x ( + C + D) x + (B C + D) x + ( + C + D) x B C + D x c copyright Hidegkuti, Powell, 0 Last revised: February, 0
16 Lecture Notes Partial Fractions page 6 Thus ( + C + D) x + (B C + D) x + ( + C + D) x B C + D x We clear the denominators by multiplication x ( + C + D) x + (B C + D) x + ( + C + D) x B C + D The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. This gives us an equation for each coe cient that forms a system of linear equations: + C + D 0 B C + D 0 + C + D 0 B C + D We will solve this system by elimination. First, we will eliminate using the rst equation. The second and fourth equations do not have in them, so there is nothing to do there. To eliminate from the third equation, we add the rst one to it. B C + D 0 C + D 0 B C + D We now have three equations with three unknowns. We will use the rst equation to eliminate B. In case of the second equation, again, there is nothing to do. We add the rst equation to the third one to eliminate B. C + D 0 C + D dding the two equations eliminates C and gives us D and so D. equation Next, we compute C using the C + D 0 C + 0 C C We can compute using the rst equation, + C + D 0 and we can compute B using the second equation, + C + D B B C + D B c copyright Hidegkuti, Powell, 0 Last revised: February, 0
17 Lecture Notes Partial Fractions page 7 Thus 0; B ; C ; and D : So our fraction, x + x x + + x can be re-written as Now we re-write the integral: x + x + +. We check: x (x + ) (x ) x + (x ) (x + ) (x + ) (x ) (x + ) (x + ) (x ) + x + (x + ) (x + ) (x + ) (x ) (x + ) (x ) x + (x ) + x + (x + ) (x + ) (x + ) (x ) x x x + x + x + x + x + x + + x + x x + x x + + x + x + x + x x x + x x x dx Thus the nal answer is x x dx x + x + + x dx arctan x ln jx + j + ln jx j + C + x x dx dx + x x + dx dx x + C arctan x ln jx + j + ln jx j + C x arctan x ln jx + j + ln jx j + C x + dx + x dx Method : The values of ; B; C; and D can be found using a slightly di erent method as follows. Consider rst the equation x + B x + + C x + + D x x We bring the fractions to the common denominator: (x + B) (x + ) (x ) (x + ) (x + ) (x ) + C x + (x ) (x + ) (x + ) (x ) + D x + (x + ) (x + ) (x + ) (x ) (x + ) (x + ) (x ) and then multiply both sides by the denominator: (x + B) (x + ) (x ) + C x + (x ) + D x + (x + ) c copyright Hidegkuti, Powell, 0 Last revised: February, 0
18 Lecture Notes Partial Fractions page 8 The equation above is about two functions; the two sides must be equal for all values of x. x into both sides: Let us substitute (x + B) (x + ) (x ) + C x + (x ) + D x + (x + ) ( ( ) + B) (( ) + ) (( ) ) + C ( ) + (( ) ) + D ( ) + (( ) + ) Let us substitute x into both sides: ( + B) (0) ( ) + C () ( ) + D () (0) (x + B) (x + ) (x ) + C x + (x ) + D x + (x + ) ( () + B) (() + ) ( ) + C + ( ) + D + ( + ) ( + B) () (0) + C () (0) + D () () D D C C Let us substitute x 0 into both sides and also C and D : (x + B) (x + ) (x ) + C x + (x ) + D x + (x + ) ( (0) + B) ((0) + ) (0 ) + C 0 + (0 ) + D 0 + (0 + ) (B) () ( ) + C () ( ) + D () () B B C + D + B + B B Let us substitute x into both sides and also B, C and D : (x + B) (x + ) (x ) + C x + (x ) + D x + (x + ) ( () + B) (() + ) ( ) + C + ( ) + D + ( + ) ( + B) () () + C () () + D () () 6 + ( + B) + C + D 6 + B + C + D c copyright Hidegkuti, Powell, 0 Last revised: February, 0
19 Lecture Notes Partial Fractions page and so 0, B, C, and D. sec x dx ln + sin x sin x + C ln jsec x + tan xj + C ln jsec x tan xj + C Solution: sec x dx cos x dx cos x cos x cos x dx Now let u sin x. Then du cos xdx. cos x sin x dx sin cos xdx x This integral can be computed by partial fractions: The left-hand side can be re-written u + B + u u cos x cos x dx u du cos x sin x dx ( u) ( + u) du u + B ( + u) B ( u) u + Bu + B ( B) u + + B + + u ( u) ( + u) ( + u) ( u) u u So we have ( B) u + + B u u We clear the denominators by multiplication ( B) u + + B The equation above is about two polynomials: they are equal to each other as functions and so they must be identical, coe cint by coe cient. This gives us an equation for each coe cient that forms a system of linear equations: we solve this system and obtain B. Now for the integral: u du + u + u B 0 + B Indeed, u + + u ( u) ( + u) u u u + + u ( u) ( + u) du du ln j uj + ln j + uj + C ln j + uj ln j uj + C ln j + sin xj ln j sin xj + C u du + + u du c copyright Hidegkuti, Powell, 0 Last revised: February, 0
20 Lecture Notes Partial Fractions page 0 Note the sign in du ln j uj + C is caused by the chain rule. slo note that the nal answer u can be re-written in several forms: ln j + sin xj ln j sin xj ln + sin x sin x ln + sin x sin x + sin x + sin x ln! s ln ( + sin x) cos x ln ( + sin x) cos ln x So, our result can also be presented as ln jsec x + tan xj + C nother form can be obtained as shown below. ln j + sin xj ln j sin xj ( + sin x) sin x + sin x cos x ln ( + sin x) cos x ln cos x + sin x cos x ln jsec x + tan xj ln + sin x sin x ln + sin x sin x sin x sin x ln sin x ( sin x) ln cos x ( sin x) cos s x ln ( sin x) ln cos x ( sin x) ln cos x sin x ln sin x cos x sin x ln cos x ln sin x cos x cos x ln jsec x tan xj So, our result can also be presented as ln jsec x tan xj + C : ctually, there are more forms possible, but 0. we will stop here. csch x dx Solution: This is an interesting application of partial fractions. csch x dx e x e x dx e x e x we now multiply both numerator and denominator by e x. e x dx e x (e x ) dx e x We proceed with a substitution: let u e x. Then du e x dx and so (e x ) (ex dx) u du dx c copyright Hidegkuti, Powell, 0 Last revised: February, 0
21 Lecture Notes Partial Fractions page This is now an integral we can easily compute via partial fractions. u u + csch x dx u du u ln je x j ln (e x + ) + C du u + u du We easily decompose u as du ln ju j ln ju + j + C u + For more documents like this, visit our page at and click on Lecture Notes. questions or comments to mhidegkuti@ccc.edu. c copyright Hidegkuti, Powell, 0 Last revised: February, 0
1.7. Partial Fractions. 1.7.1. Rational Functions and Partial Fractions. A rational function is a quotient of two polynomials: R(x) = P (x) Q(x).
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