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Integration-By-Parts.Pdf Integration INTEGRATION BY PARTS Graham S McDonald A self-contained Tutorial Module for learning the technique of integration by parts ● Table of contents ● Begin Tutorial c 2003 [email protected] Table of contents 1. Theory 2. Usage 3. Exercises 4. Final solutions 5. Standard integrals 6. Tips on using solutions 7. Alternative notation Full worked solutions Section 1: Theory 3 1. Theory To differentiate a product of two functions of x, one uses the product rule: d dv du (uv) = u + v dx dx dx where u = u (x) and v = v (x) are two functions of x. A slight rearrangement of the product rule gives dv d du u = (uv) − v dx dx dx Now, integrating both sides with respect to x results in Z dv Z du u dx = uv − v dx dx dx This gives us a rule for integration, called INTEGRATION BY PARTS, that allows us to integrate many products of functions of x. We take one factor in this product to be u (this also appears on du the right-hand-side, along with dx ). The other factor is taken to dv be dx (on the right-hand-side only v appears – i.e. the other factor integrated with respect to x). Toc JJ II J I Back Section 2: Usage 4 2. Usage We highlight here four different types of products for which integration by parts can be used (as well as which factor to label u and which one dv to label dx ). These are: sin bx (i) R xn· or dx (ii) R xn·eaxdx cos bx ↑ ↑ ↑ ↑ dv dv u dx u dx sin bx (iii) R xr· ln (ax) dx (iv) R eax · or dx cos bx ↑ ↑ ↑ ↑ dv dv dx u u dx where a, b and r are given constants and n is a positive integer. Toc JJ II J I Back Section 3: Exercises 5 3. Exercises Click on Exercise links for full worked solutions (there are 14 exer- cises in total) Z Exercise 1. x cos x dx Z Exercise 2. x2 sin x dx Z Exercise 3. xexdx Z Exercise 4. x2e4xdx Z Exercise 5. x2 ln x dx ● Theory ● Integrals ● Final solutions ● Tips ● Notation Toc JJ II J I Back Section 3: Exercises 6 Z Exercise 6. (x + 1)2 ln 3x dx Z Exercise 7. e2x cos x dx Z Exercise 8. e−x sin 4x dx 1/2 Z Exercise 9. xe2x dx 0 π/4 Z Exercise 10. x sin 2x dx 0 ● Theory ● Integrals ● Final solutions ● Tips ● Notation Toc JJ II J I Back Section 3: Exercises 7 1 Z Exercise 11. x4 ln 2x dx 1/2 π Z x Exercise 12. 3x2 cos dx 2 0 Z Exercise 13. x3ex dx Z Exercise 14. e3x cos x dx ● Theory ● Integrals ● Final solutions ● Tips ● Notation Toc JJ II J I Back Section 4: Final solutions 8 4. Final solutions 1. x sin x + cos x + C , 2. −x2 cos x + 2x sin x + 2 cos x + C , 3. (x − 1) ex + C , 1 4x 2 4. 32 e 8x − 4x + 1 + C , 1 3 5. 9 x (3 ln x − 1) + C , 1 3 1 3 1 2 1 6. 3 (x + 1) ln 3x − 9 x − 2 x − x − 3 ln x + C , 1 2x 7. 5 e (sin x + 2 cos x) + C , Toc JJ II J I Back Section 4: Final solutions 9 1 −x 8. − 17 e (4 cos 4x − sin 4x) + C , 1 9. 4 , 1 10. 4 , 1 31 11. 5 ln 2 − 800 , 12. 6 π2 − 8 , 13. ex x3 − 3x2 + 6x − 6 + C , 1 3x 14. 10 e (sin x + 3 cos x) + C . Toc JJ II J I Back Section 5: Standard integrals 10 5. Standard integrals f (x) R f(x)dx f (x) R f(x)dx n+1 n xn+1 n 0 [g(x)] x n+1 (n 6= −1) [g (x)] g (x) n+1 (n 6= −1) 1 g0(x) x ln |x| g(x) ln |g (x)| x x x ax e e a ln a (a > 0) sin x − cos x sinh x cosh x cos x sin x cosh x sinh x tan x − ln |cos x| tanh x ln cosh x x x cosec x ln tan 2 cosech x ln tanh 2 sec x ln |sec x + tan x| sech x 2 tan−1 ex sec2 x tan x sech2 x tanh x cot x ln |sin x| coth x ln |sinh x| 2 x sin 2x 2 sinh 2x x sin x 2 − 4 sinh x 4 − 2 2 x sin 2x 2 sinh 2x x cos x 2 + 4 cosh x 4 + 2 Toc JJ II J I Back Section 5: Standard integrals 11 f (x) R f (x) dx f (x) R f (x) dx 1 1 tan−1 x 1 1 ln a+x (0<|x|<a) a2+x2 a a a2−x2 2a a−x (a > 0) 1 1 ln x−a (|x| > a>0) x2−a2 2a x+a √ 2 2 √ 1 sin−1 x √ 1 ln x+ a +x (a > 0) a2−x2 a a2+x2 a √ 2 2 (−a < x < a) √ 1 ln x+ x −a (x>a>0) x2−a2 a √ √ 2 √ 2 h i 2 2 a −1 x 2 2 a −1 x x a2+x2 a − x 2 sin a a +x 2 sinh a + a2 √ √ i √ 2 h i x a2−x2 2 2 a −1 x x x2−a2 + a2 x −a 2 − cosh a + a2 Toc JJ II J I Back Section 6: Tips on using solutions 12 6. Tips on using solutions ● When looking at the THEORY, INTEGRALS, FINAL SOLU- TIONS, TIPS or NOTATION pages, use the Back button (at the bottom of the page) to return to the exercises ● Use the solutions intelligently. For example, they can help you get started on an exercise, or they can allow you to check whether your intermediate results are correct ● Try to make less use of the full solutions as you work your way through the Tutorial Toc JJ II J I Back Section 7: Alternative notation 13 7. Alternative notation In this Tutorial, we express the rule for integration by parts using the formula: Z dv Z du u dx = uv − v dx dx dx But you may also see other forms of the formula, such as: Z Z dg f(x)g(x)dx = F (x)g(x) − F (x) dx dx where dF = f(x) dx Of course, this is simply different notation for the same rule. To see this, make the identifications: u = g(x) and v = F (x). Toc JJ II J I Back Solutions to exercises 14 Full worked solutions Exercise 1. We evaluate by integration by parts: Z Z x cos x dx = x · sin x− (1) · sin x dx, i.e. take u = x du giving = 1 (by differentiation) dx dv and take = cos x dx giving v = sin x (by integration), Z = x sin x − sin x dx = x sin x − (− cos x) + C, where C is an arbitrary = x sin x + cos x + C constant of integration. Return to Exercise 1 Toc JJ II J I Back Solutions to exercises 15 Exercise 2. Z Z x2 sin x dx = x2 · (− cos x) − (2x) · (− cos x) dx , i.e. take u = x2 du giving = 2x dx dv and take = sin x dx giving v = − cos x , Z = −x2 cos x + 2 x cos x dx | {z } ,→ we need to use integration by parts again! Toc JJ II J I Back Solutions to exercises 16 Z = −x2 cos x + 2 x sin x − (1) · sin x dx , as in question 1, Z = −x2 cos x + 2x sin x − 2 sin x dx = −x2 cos x + 2x sin x − 2 · (− cos x) + C = −x2 cos x + 2x sin x + 2 cos x + C. Return to Exercise 2 Toc JJ II J I Back Solutions to exercises 17 Exercise 3. Z Z x exdx = x · ex − (1) · exdx, i.e. take u = x du giving = 1 dx dv and take = ex dx giving v = ex, Z = x ex − exdx = x ex − ex + C = (x − 1) ex + C. Return to Exercise 3 Toc JJ II J I Back Solutions to exercises 18 Exercise 4. Z 1 Z 1 x2 e4xdx = x2 · e4x − 2x · e4xdx , 4 4 i.e. take u = x2 du = 2x dx dv = e4x dx 1 v = e4x , 4 1 1 Z = x2e4x − x e4xdx 4 2 | {z } ,→ now use integration by parts again Toc JJ II J I Back Solutions to exercises 19 1 1 1 Z 1 = x2e4x − x · e4x − 1 · e4xdx , 4 2 4 4 i.e. this time u = x du = 1 dx dv = e4x dx 1 v = e4x, 4 1 1 1 1 Z = x2e4x − xe4x − · − · e4xdx 4 8 2 4 1 1 1 Z = x2e4x − xe4x + e4xdx 4 8 8 1 1 1 1 = x2e4x − xe4x + · e4x + C 4 8 8 4 Toc JJ II J I Back Solutions to exercises 20 8 4 1 = x2e4x − xe4x + e4x + C 32 32 32 1 = e4x 8x2 − 4x + 1 + C. 32 Return to Exercise 4 Toc JJ II J I Back Solutions to exercises 21 Exercise 5. Z 1 Z 1 1 x2 ln x dx = (ln x) · x3 − · x3 dx , i.e. u = ln x 3 x 3 du 1 = dx x dv = x2 dx 1 v = x3 , 3 1 1 Z = x3 ln x − x2dx 3 3 1 1 1 = x3 ln x − · x3 + C 3 3 3 1 1 = x3 ln x − x3 + C 3 9 1 = x3 (3 ln x − 1) + C.
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