Integration by Parts

1 Joe FosterIntegration by PartsTo reverse the chain rule we have the method ofu-substitution. To reverse the product rule we also have a method, calledIntegration by Parts . The formula is given by:Theorem( Integration by Parts Formula) f(x)g(x)dx=F(x)g(x) F(x)g (x)dxwhereF(x) is an anti-derivative off(x).Remember, all of the techniques that we talk about are supposed to make integrating easier! Even though this formulaexpresses one integral in terms of a second integral, the idea is that the second integral, F(x)g (x)dx, is easier to key to Integration by Parts is making the right choice forf(x) andg(x). Sometimes we may need to try multipleoptions before we can apply the formula. Let s see it in 1 Find xcos(x) have to decide what to assign tof(x) and what to assign tog(x). Our goal is to make the integraleasier. One thingto bear in mind is that whichever term we let equalg(x) we need to differentiate - so if differentiating makes a part of theintegrand simpler that s probably what we want!

2 In this cases differentiating cos(x) gives sin(x), which is no easier todeal with. But differentiatingxgives 1 whichissimpler. So we have,g(x) =xf(x) = cos(x)g (x) = 1F(x) = sin(x) xcos(x)dx=xsin(x) sin(x)dx=xsin(x) + cos(x) +CExample 2 Evaluate 40xe (x) =xf(x) =e xg (x) = 1F(x) = e x 40xe xdx= xe x 40 40 e xdx= xe x e x 40=[ 4e 4 e 4] [0 e 0]= 5e 4+ 1= 1 5e 4 Page 1 of 8 MATH 142 - Integration by PartsJoe FosterExample 3 Evaluate (x) =x2f(x) =exg (x) = 2xF(x) =ex x2exdx=x2ex 2 s at this point we see that we still cannot integrate the integral on the write easily. This is we mayhave to apply the Integration by Parts formula more than once!g1(x) =xf1(x) =exg 1(x) = 1F1(x) =ex x2exdx=x2ex 2 xexdx=x2ex 2[xex exdx]=x2ex 2xex+ 2ex+C=(x2 2x+ 2)ex+CThe previous technique works for any integral of the form xnemxdx, wherenis anypositiveinteger andmis any ifnwas negative? Then this case we would setg(x) = 4In Example 3 we have to apply the Integration by Parts Formulamultiple times.

3 There is a convenient wayto book-keep our work. This is done by creating a table. Let s see how by examining Example 3 again. (x) =x2andf(x) =ex. Then,Differentiateg(x)Integratef(x)x2ex2 xex2ex0ex+ +Then the integral is, x2exdx= +x2 ex 2x ex+ 2 ex+C=(x2 2x+ 2)ex+CWehaveactually used the Integration by Parts formula, but we have just made our lives easier by condensing the workinto a neat table. This method is extremely useful when Integration by Parts needs to be used over and over 2 of 8 MATH 142 - Integration by PartsJoe FosterThe next example exposes a potential flaw in always using the tabular method above. Sometimes applying the integrationby Parts formula may never terminate, thus your table will get awfully 5 Find the integral exsin(x) need to apply Integration byParts twice before we see something:(1)u=exdv= sin(x)du=exdxv= cos(x)(2)u=exdv= cos(x)du=exdxv= sin(x) exsin(x)dx= excos(x) + excos(x)dx= excos(x) +(exsin(x) exsin(x)dx)= excos(x) +exsin(x) exsin(x)dxNotice that now the integral we are interested in, exsin(x)dx, appears on both theleft and right hand side of the equation.

4 So, if we add this integral to both sides weget= 2 exsin(x)dx=ex( cos(x) + sin(x))= exsin(x)dx=ex(sin(x) cos(x))2 This trick comes up often when we are dealing with the product of two functions with non-terminating this we mean that you can keep differentiating functions likeexand trig functions indefinitely and never reach0. Polynomials on the other hand will eventually terminate and theirnthderivative (wherenis the degree of thepolynomial) is identically 3 of 8 MATH 142 - Integration by PartsJoe FosterPractice ProblemsTry some of the problems below. If you get stuck, don t worry!There are hints on the next page! But do try withoutlooking at them first, chances are you won t get hints on your tsin(2t)dt2. x2cos(3x)dx3. sin 1(x)dx4. p5ln(p)dp5. 10(x2+ 1)e xdx6. 94ln(y) ydy7. 0x3cos(x)dx8. 31tan 1(1/x)dx9. 21(ln(x))2x3dx10. (ln(x))2dx11. 4xcos(2 3x)dx12. 06(2 + 5x)ex/3dx13. (t2+ 3t) sin(2t)dt14. 0x2cos(4x)dx15. (4z3 9z2+ 7z+ 3)e zdz16. 8te7tdt17. x3ln(3 x)dx18.

5 Tsec2(2t)dt19. e cos(2 )d 20. e2zcos(z/4)dz21. 21ln(x)x2dx22. 10ye2ydy23. 1/20cos(x) ln(sin(x))dx24. x4(ln(x))2dxChallenge Problems1. ln(x)dx2. t7sin(2t4)dt3. 41(2 x)2ln(4x)dx4. tan 1(x)dx5. sin 1(x)dx6. cos( x)dx7. t3e t2dt8. xln(1 +x)dx9. sin(ln(x))dxPage 4 of 8 MATH 142 - Integration by PartsJoe FosterHints to Practice (t) =t2. Apply twice,start withg(x) = (x) = sin 1(x) (p) = ln(p)5. Apply twice,start withg(x) =x2+ (y) = ln(y)7. Apply three times,start withg(x) = (x) = tan 1(1/x)9. Apply twice,start withg(x) = (ln(x)) (x) = (ln(x)) (x) = (x) = 2 + 5x13. Apply twice,start withg(t) =t2+ 3t14. Apply twice,start withg(x) =x215. Apply three times,start withg(z) = 4z3 9z2+7z+ (t) = (x) = ln(x) (t) =t19. Think Example Think Example (x) = ln(x) (y) = (x) = ln(sin(x))24. Apply twice,start withg(x) = (ln(x))2 Hints to Challenge (x) = ln(x),f(x) = (t) = (x) = ln(4x) (x) = tan 1(x) (x) = sin 1(x) x, then t2, then 1 +x, then , then 5 of 8 MATH 142 - Integration by PartsJoe FosterAnswers to Practice (sin(2t) 2tcos(2t)) + ((9x2 2) sin(3x) + 6xcos(3x))+ 1(x) + 1 x2+ (6 ln(p) 1) + 6e6.

6 4(ln(274) 1)7. 3(4 2) ((2 3 3) + ln(64)) (3 ln(2)(ln(2) + 1)) (ln(x)2 2 ln(x) + 2)+ (4 cos(2 3x) 12xsin(2 3x)) +C12. 39 (( 2t2 6t+ 1)cos(2t) + (2t+ 3) sin(2t))+C14. 815. (4z3+ 3z+13z+ 16)e z+ (56t 8) + (5 ln(x) 2) + (2ttan(2t) + ln|cos(2t|) + (8 cos(z4)+ sin(z4))+ ln(2) (1 3e2)23. sin(12) (ln(sin(12)) 1) (25 ln(x)2 10 ln(x) + 2)+CAnswers to Challenge (ln(x) 1) + (sin(2t4) 2t4cos(2t4))+C3. ln(65,536 22/3) 1(x) ln( 1 +x2) + 1(x) + 1 x2+C6. 2 xsin( x) + 2 cos( x) +C7. 12et2(t2+ 1)+ (2(x2 1) ln(1 +x) x+ 2)+ (sin(ln(x)) cos(ln(x))) +CPage 6 of 8 MATH 142 - Integration by PartsJoe FosterSolutions to Practice Problems1. Letg(t) =tandf(t) = sin(2t). Sog (t) = 1 andF(t) = 12cos(2t). Then, tsin(2t)dt= 12tcos(2t) +12 cos(2t)dt= 12tcos(2t) +14sin(2t) +C=14(sin(2t) 2tcos(2t)) +C2. Letg(x) =x2andf(x) = cos(3x). Then,Differentiateg(x)Integratef(x)x2cos (3x)2x13sin(3x)2 19cos(3x)0 127sin(3x)+ + x2cos(3x)dx=13x2sin(3x) +29xcos(3x) 227sin(3x) +C=127((9x2 2)sin(3x) + 6xcos(3x))+C3. Letg(x) = sin 1(x) andf(x) = 1.)

7 Sog (x) =1 1 x2andF(x) = 1 x2. So 12du=x dx. Then, sin 1(x)dx=xsin 1(x) x 1 x2dx=xsin 1(x) +12 1 udu=xsin 1(x) + u+C=xsin 1(x) + 1 x2+C4. Letg(p) = ln(p) andf(p) =p5. Sog (p) =1pandF(p) =16p6. Then, p5ln(p)dp=16p6ln(p) 16 p5dp=16p6ln(p) 136p6+C=136p6(6 ln(p) 1) +C5. Letg(x) =x2+ 1 andf(x) =e x. Then,Differentiateg(x)Integratef(x)x2+ 1e x2x e x2e x0 e x+ + 10(x2+ 1)e xdx= (x2+ 1)e x 2xe x 2e x 10= e x(x2+ 2x+ 3) 10= 6e 1+ 3=3e 6e6. Letg(y) = ln(y) andf(y) =y 1/2. Sog (y) =y 1andF(y) = 2y1/2. Then, 94ln(y) ydy= 2 yln(y) 94 2 94y 1/2dy= 2 yln(y) 4 y 94= 6 ln(9) 12 [4 ln(4) 8]= 4 ln(27) 4 4 ln(4) = 4(ln(274) 1)Page 7 of 8 MATH 142 - Integration by PartsJoe Foster7. Letg(x) =x3andf(x) = cos(x). Then,Differentiateg(x)Integratef(x)x3cos (x)3x2sin(x)6x cos(x)6 sin(x)0cos(x)+ + 0x3cos(x)dx=x3sin(x) + 3x2cos(x) 6xsin(x) 6 cos(x) 0=x(x2 6)sin(x) + 3(x2 2)cos(x) 0= 0 3( 2 2) [0 6]= 3(4 2)8. Letg(x) = tan 1(1/x) andf(x) = 1. Sog (x) = 1x2+ 1andF(x) = + 1. So12du=x dx. Then, 31tan 1(1x)dx=xtan 1(x)(1x) 31+ 31xx2+ 1dx=xtan 1(1x) 31+12 x= 3x=11udu=xtan 1(1x)+12ln|u| x= 3x=1=xtan 1(1x)+12ln(x2+ 1) 31= 3 tan 1( 33)+12ln(4) [tan 1(1) +12ln(2)]= 3 6+ ln(2) 4 12ln(2)=112((2 3 3) + ln(64))9.

8 Letg1(x) = (ln(x))2andf1(x) =1x3. Sog 1(x) =2 ln(x)xandF1(x) = (x) = ln(x) andf2(x) =1x3. Sog 2(x) =1xandF2(x) = 12x2. Then, (ln(x))2xdx= (ln(x))22x2+ ln(x)x3dx= (ln(x))22x2 ln(x)2x2+12 1x3dx= (ln(x))2+ ln(x)2x2 14x2+C. 21(ln(x))2xdx= 2(ln(x))2+ 2 ln(x) + 14x2 21= 2(ln(2))2+ 2 ln(2) + 116 [ 0 + 0 + 14]=116(3 2(ln(2))2 2 ln(2))10. Letg1(x) = (ln(x))2andf1(x) = 1. Sog 1(x) =2 ln(x)xandF1(x) = (x) = ln(x) andf2(x) = 1. Sog 2(x) =1xandF2(x) =x. Then, (ln(x))2dx=x(ln(x))2 2 ln(x)dx=x(ln(x))2 2[xln(x) 1dx]=x(ln(x))2 2 [xln(x) x] +C=x((ln(x))2 2 ln(x) + 2)+CPage 8 of 8