Transcription of 25Integration by Parts
1 Page 1 of 7 3 Formula =vduuvudv I. Guidelines for Selecting u and dv: (There are always exceptions, but these are generally helpful.) L-I-A-T-E Choose u to be the function that comes first in this list: L: Logrithmic Function I: Inverse Trig Function A: Algebraic Function T: Trig Function E: Exponential Function Example A: dxxxln3 *Since lnx is a logarithmic function and 3xis an algebraic function, let: u = lnx (L comes before A in LIATE) dv = 3xdx du = x1 dx v = =443xdxx =vduuvxdxxln3 dxxxxx144)(ln44 = dxxxx =34414)(ln Cxxx+ =441)(ln444 Cxxx+ =16)(ln444 ANSWER Integration By Parts Page 2 of 7 Example B: dxxx)ln(cossin u = ln(cosx) (Logarithmic Function) dv = sinx dx (Trig Function [L comes before T in LIATE]) du = dxxdxxxtan)sin(cos1 = v = =xdxxcossin =vduuvdxxx)ln(cossin =dxxxxx)tan)(cos()cos))((ln(cos =dxxxxxxcossin)(cos)ln(coscos =dxxxxsin)ln(coscos Cxxx++ =cos)ln(coscos ANSWER Example C: dxx1sin *At first it appears that integration by Parts does not apply, but let: xu1sin = (Inverse Trig Function) dxdv1= (Algebraic Function) dxxdu211 = ==xdxv1 = vduuvdxx1sin dxxxxx = 2111))((sin dxxxxx = )2()1(21sin2/121 Cxxx+ += )2()1(21sin2/121 Cxxx+ += 211sin ANSWER Page 3 of 7 II.
2 Alternative General Guidelines for Choosing u and dv: A. Let dv be the most complicated portion of the integrand that can be easily integrated. B. Let u be that portion of the integrand whose derivative du is a simpler function than u itself. Example: dxxx234 *Since both of these are algebraic functions, the LIATE Rule of Thumb is not helpful. Applying part (A) of the alternative guidelines above, we see that 24xx is the most complicated part of the integrand that can easily be integrated. Therefore: dxxxdv24 = 2xu= (remaining factor in integrand) dxxdu2= = =dxxxdxxxv2/122)4)(2(214 2/322/32)4(31)4(3221xx = = = vduuvdxxx334 =dxxxxx)2()4(31)4(31)(2/322/322 =dxxxxx)2()4(31)4(32/322/322 Cxxx+ =52)4(31)4(32/522/322 Cxxx+ =2/522/322)4(152)4(3 Answer Page 4 of 7 III.
3 Using repeated Applications of Integration by Parts : Sometimes integration by Parts must be repeated to obtain an answer. Example: dxxxsin2 2xu= (Algebraic Function) dxxdvsin= (Trig Function) dxxdu2= ==xdxxvcossin =vduuvdxxxsin2 =dxxxxx2cos)cos(2 + =dxxxxxcos2cos2 Second application of integration by Parts : xu= (Algebraic function) (Making same choices for u and dv) xdvcos= (Trig function) dxdu= ==xdxxvsincos + =][2cossin22vduuvxxdxxx + =]sinsin[2cos2dxxxxxx ]cossin[2cos2cxxxxx+++ = cxxxxx+++ =cos2sin2cos2 Answer Note: DO NOT switch choices for u and dv in successive applications. Page 5 of 7 Example: dxxexcos xucos= (Trig function) dxedvx= (Exponential function) dxxdusin = ==xxedxev =vduuvdxxexcos =dxxeexxx)sin(cos +=dxxeexxxsincos Second application of integration by Parts : xusin= (Trig function) (Making same choices for u and dv) dxedvx= (Exponential function) dxxducos= ==xxedxev +=)(coscosvduuvxedxxexx +=dxxeexxedxxexxxxcossincoscos Note appearance of original integral on right side of equation.
4 Move to left side and solve for integral as follows: Cxexedxxexxx++= sincoscos2 Cxexedxxexxx++= )sincos(21cos Answer Note: After each application of integration by Parts , watch for the appearance of a constant multiple of the original integral. Page 6 of 7 Practice Problems: 1. dxexx3 2. dxxx2ln 3. dxxxcos2 4. dxxxxcossin 5. dxx1cos 6. dxx2)(ln 7. dxxx239 8. dxxexsin2 9. dxxx12 10. dxxx3)(ln1 Page 7 of 7 Solutions: 1. Cexexx+ 33 xu3= dxedvx = 2. Cxxx+ 1ln xuln= dxxdv21= 3. Cxxxxx+ +sin2cos2sin2 2xu= dxxdvcos= 4. Cxxx++ 82sin42cos note: xxxcossin22sin= xu= dxxxdvcos2sin= 5. Cxxx+ 211cos xu1cos = dxdv= 6. Cxxxxx++ 2ln2)(ln2 2)(lnxu= dxdv= 7. Cxxx+ 2/522/322)9(152)9(3 2xu= dxxxdv2/12)4( = 8. Cxexexx+ 5cos5sin222 xusin= dxedvx2= 9.
5 Cxxxxx+ + 105)1(1615)1(83)1(22/72/52/32 2xu= dxxdv2/1)1( = 10. Cx+ 2)(ln21 33)(ln)(ln1 ==xxu dxxdv1=
