Integrale rezolvate

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1. x 2 x 2 1 dx =? Solutie : x 2 1 1 x 2 1 dx = x arctg x C 2. x 3 x 2 1 dx =? Solutie : x 3 dx x 2 dx dx = x 4 4 x 3 3 x C 3. 1 2x 1 dx =? Solutie : Observam ca ˙ ln 2x 1 4. 1 4x 5 dx =? Solutie : Se rezolvă în mod similar cu cea de mai sus numai ca , vom pune 1 4 în faţa integralei deoarece 1 4x 5 = ln 4x 5 ' 1 4x 5 dx = 1 4 ln 4x 5 ' dx = 1 4 ln 4x 5 5. 2x 2x 2 3 dx =? Solutie : ! De obicei când întâlnim radicalul la numitor derivam si observam ce forma obtinem : Pentru cazul nostru observam ca : 2x 2 3 ' = 4x 2 2x 2 3 = 2x 2x 2 3 ceea ce reprezinta exact valoarea din integrală 2x 2x 2 3 dx = 2x 2 3 ' dx = 2x 2 36. x 5x 2 2 dx =? Solutie : 5x 2 2 ' = 10x 2 5x 2 2 = 5x 5x 2 2 rezulta x 5x 2 2 = 5x 2 2 5 ' x 5x 2 2 dx = 1 5 5x 2 2 ' dx = 1 5 5x 2 2

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Transcript of Integrale rezolvate

1.( x2( x2+1))dx=?Solutie :(( x2+11)( x2+1) ) dx=xarctg( x)+C2.( x3+x2+1) dx=?Solutie :( x3) dx+( x2)dx+dx= x44 +x33 +x+C3. ( 1( 2x+1)) dx=?Solutie :Observam ca ln( 2x+1)4.( 14x+5)dx=?Solutie :Se rezolvn mod similar cu cea de mai sus numai ca , vom pune14 n faaintegralei deoarece1(4x+5)=(ln( 4x+5))'( 1( 4x+5)) dx=14(ln (4x+5)) ' dx=14 ln( 4x+5)+5.(2x.(2x2+3))dx=?Solutie :! De obicei cnd ntlnimradicalul la numitor derivamsi observamce forma obtinem:Pentrucazul nostru observam ca :(.2x2+3)' =4x2.2x2+3=2x.2x2+3ceea ce reprezinta exact valoarea din integral2x. 2x2+3 dx=(.2x2+3)' dx=.2x2+3+6.x.5x2+2 dx=?Solutie :(.5x2+2)' =10x2.5x2+2=5x.5x2+2 rezulta x.5x2+2=(.5x2+25 )'x.5x2+2 dx=15(.5x2+2)' dx=15 .5x2+2+ 8.cos(3x)dx=?Solutie:Daca derivam , (cos(3x)) ' =3sin(3x)Dar (sin (3x)) ' =3cos (3x) rezulta cos(3x)=(sin(3x)3 )'Decicos(3x) dx=13(sin(3x)) ' dx=13 sin(3x)+9. I =( x2+2x+1 x ) dx , x0; I =?Solutie :I =x2 dx+2x dx+1 x dxI =x3 3 +x2 +ln (x)+!Observatie: Rezultatul contine ln(x) pentruc din ipotez timc x0.10. I =( x+1 x ) dx , x>0; I =?Solutie :I =x dx+dxx =x2 2 +ln ( x)+!Observatie: n acest caz rezultatul conineln ( x) pentru ca x0.11.I=x3x5 dx , x>0; I =?Solutie :I =(xx5+3x5)dx=dxx4 +3dxx5I =x4dx+3x5dx=x4+14+1+3x5+15+1 +I =13x3 34x4+12. I =(asin( x)+bcos( x))dx ; a , b; I =?Solutie :I =asin( x) dx+bcos( x)dx=acos( x)+bsin( x)+13. I =cos(2x)sin2( x) cos2( x) dx , x(0 ,n2 ) ; I =?Solutie :Scriem cos(2x)=cos2( x)sin2( x) i obinem:I =cos2( x)sin2( x)sin2( x) cos2( x) dx=(1sin2( x)1cos2( x) )dxI =dxsin2( x)dxcos2( x)=ctg ( x)tg ( x)+14. I =dx.14x2 , x(12 ,12) ; I =?Solutie :I =dx.12(2x)2=12 arcsin (2x)+Verificare :(12 arcsin(2x)) ' =12 1.12(2x)2 2=1.12(2x)215. I =(2sin2( x)+1cos2( x))dx , x(0,n2 ) ; I =?Solutie :I =2dxsin2( x)+dxcos2( x)=2ctg( x)+tg ( x)+16. I =dx.169x2 , x(43 ,43) ; I =?Solutie :I se mai pote scrie i astfel :I =dx.42(3x)2 =13 arcsin(3x4 )+Verificare :(13 arcsin (3x4 ))' =13 1.42(3x)2 3=1.169x217. I =dx.4x2 , x(2,2) ; I =?Solutie :I =dx.22x2=arcsin(x2 )+18. I =dxx2+4 ; I =?Solutie:I =dxx2+22=12 artcg(x2)+19. I =dx4x2+1 : I =?Solutie:I =dx( 2x)2+12=12 arctg (2x)+20. I =(.x+3. x+4.x) dx , x0 ; I =?Solutie:I =x12dx+x13dx+x14dxI =x12+112+1+x13+113+1+x14+114+1I =x3232+x4343+x5454+I =23 .x3+34 3.x4+45 4.x5+I =23 x.x +34 x 3. x +45 x 4.x +21.I=(2. x33. x )dx , x>0; I =?Solutie :I =2x12dx3x13dx= 2x12+112+1 3x13+113+1+I =2. x12 3x2323 +I =4 . x92 3. x2+22. I =( 2x+ex) dx , x; I =?Solutie :I =2xdx+exdx=2xln2+ex+! Amobservat c (2x)' =2xln2 , deci 2x=(2x)'ln223. I =(2ex3x)dx , x; I =?Solutie:I =2exdx3xdx=2ex3xln3+Verificare :( 2ex3x) ' =2ex3xln3ln3 =2ex3x24. I =dxx21 , x(1,1); I =?Solutie:I =1x21 dx=12 lnx1x+1+=ln.x1x+1+25. I =dxex , x; I =?Solutie :I =exdx=ex+26. I =( x21)2x4 dx , x>0; I =?Solutie :( x21)2=x42x2+1I =x4x4 dx2x2x4 dx+1x4 dx=1dx21x2 dx+1x4 dxI =dx+2x2dx+x4dx=x+2x 13x3+27. I =1.1x21x2 dx , x(1,1); I =?Solutie:I =(11x2.1x21x2 )dx=dxx21dx.1x2I =12 lnx1x+1arcsin( x)+Dar , innd cont c x(1,1) , I va fi :I =12 ln (x1x+1)arcsin( x)+28. I =3+. x2+4x2+4 dx , x; I =?Solutie :I =(3x2+4+. x2+4x2+4 )dx =3dxx2+4+dx. x2+4I =32 arctg(x2 )+ln ( x+. x2+4)+29. I =cos2( x)cos4( x) dx , x(0,n2 ) ; I =?Solutie:I =dxcos2( x)=tg ( x)+30. I =dx. x2+25 , x; I =?Solutie:I =dx.x2+52=lnx+. x2+52+1Integrarea prin pri!Nu din pri :DFormula:f g ' dx=f gf 'g dxS se calculeze integralele:1.lnx dx , x>0Solutie :Alegem f ( x)=ln ( x) , g ' ( x)=1. De aici :f ' ( x)=1, g( x)=xFolosind formula integrrii prin pri , obinem:x ln ( x) dx=x' ln( x) dx=xln( x)x1x dx==xln( x)x+2.xln( x) dx , x>0Soltuie :Alegem f ( x)=ln( x) , g ' ( x)=x. nconcluzie :f ' ( x)=1x , g ( x)=x22Aplicm formula integrrii prin pri :xln( x) dx=ln ( x)(x22 )' dx=ln( x)x22 12x21x dx==x22 ln ( x)14 x2+3.ln2( x)dx , x>0Solutie :Notm f ( x)=ln2( x) , g' ( x)=1.Deci :f ' ( x)=2x ln( x) , g( x)=xGsim:ln2( x) dx=x' ln ( x) dx=xln2( x)2ln( x)x xdx==xln2( x)2ln( x) dx Folosind ex1. obinem:ln2( x)dx=xln2( x)2( xln( x)x)+==x(ln2( x)2ln ( x)+2)+14.x2ln( x) dx , x>0Solutie :f ( x)=ln ( x) , g ' ( x)=x2si avem:f ' ( x)=1x , g ( x)=x33Aplicnd formulaobinem:x2ln ( x)dx=(x33 )' ln( x)13x31x dx=x33 ln( x)13x33 +==x33 ln ( x)19 x3+5.ln ( x)x dx , x>0Solutie :f ( x)=ln ( x) , g' ( x)=1xf ' ( x)=1x , g ( x)=ln( x)Aplicm formula:ln ( x)x dx=( ln( x)) 'ln( x) dx=ln2( x)1x ln ( x) dxObservmcln( x)x dx=ln2( x)ln ( x)x dx , deci2ln( x)x dx=ln2( x)+, n final :ln ( x)x dx=12 ln2( x)+6.x2exdx , xSolutie :f ( x)=ex, g ' ( x)=x2 , atunci :f ' ( x)=ex, g( x)=x3 3 , deci :x2exdx=(x33 ) 'exdx=x3 3 ex1 3 x3exdxObservm c integrala astfel obinut este mult mai complicatAtunci vomalege f ( x)=x2, g ' ( x)=excuf ' ( x)=2x , g( x)=exDeci :x2exdx=x2( ex)' dx==x2ex2 xexdxAplicmncodat formuladeintegrare prin pri i alegem:f ( x)=x , g' ( x)=exastfel nct :f ' ( x)=1, g( x)=exsi obinem:xexdx=x( ex)' dx=xexexx' dx= xexex+n final :x2exdx=x2ex2( xexex)+==ex( x22 x+2)+7.( x22x1) exdx , xSolutie :Considerm f ( x)=x22x1si g' ( x)=excuf ' ( x)=2x2 si g( x)=exAplicnd formula obinem:( x22x1) exdx=( x22x1)(ex)' dx=newkine =( x22x1)ex2( x1) exdxLund separat :( x1)exdx=xexdxexdx=(conformex6)==xexex+n final :( x22x1) exdx=( x22x1) ex2xex+4 ex+==ex( x24x+3)+8.x sin( x) dx , xSolutie :Notm f ( x)=x , g ' ( x)=sin( x) si avem:f ' ( x)=1, g ( x)=cos ( x)Deci :xsin( x) dx=x(cos ( x))' dx==xcos( x)cos( x) dx==xcos( x)+sin( x)+9.x2sin( x) dx , xSolutie :f ( x)=x2, g ' ( x)=sin( x) -- f ' ( x)=2x , g ( x)=cos( x) , integrala devine :x2sin( x) dx=x2(cos ( x)' ) dx==x2cos ( x)2xcos( x) dx , notam 2xcos( x)dx=I 'I ' =2xcos( x) dx=2int x(sin( x)) ' dx==2xsin ( x)2x(sin( x))' dx==2x sin( x)+2cos( x)+Finalizare :x2sin( x) dx=x2cos( x)+2xsin( x)+2cos( x)+10.sin2( x) dx , xSolutie :Lum f ( x)=sin2( x) si g' ( x)=1-- f ' ( x)=2sin( x)cos ( x)=sin( 2x) si g( x)=xsin2( x)dx=( x)' sin2( x) dx=xsin2( x)xsin(2x)dx notam xsin(2x)dx=I 'I ' =12x(cos (2x))' dx=12 xcos( 2x)12cos( 2x) dx==12 xcos(2x)12 sin( 2x)12+Finalizare :sin2( x)dx=x(sin2( x)cos(2x)2 )14 sin( 2x)+11.exsin( x) dx , xSolutie:Notm f ( x)=ex, g ' ( x)=sin( x) -- f ' ( x)=ex, g ( x)=cos( x)n concluzie:I =exsin( x) dx=ex(cos ( x)) dx==excos( x)+excos ( x) dx notam excos( x) dx=I 'I ' =ex(sin( x))' dx=exsin( x)exsin( x) dx dar exsin( x) dx=IDeci :I =excos ( x)+exsin( x)I +I =12 ex(sin( x)cos ( x))+Obs: I'->citim I prim i nu I derivat ->l-am ales ca pe o notaie``12.. x29dx , x>3Solutie :I = . x291. x29dx=(amraionalizat )=x29. x29 dx==x2. x29 dx_I19dx.x29_I2unde I =I1I2I2=9lnx+.x29Pentru a calcula I1,notm f ( x)=x , g ' ( x)=(. x29) ' adic g ' ( x)=2 x2.x29=xx29 unde:f ' ( x)=1 si g ( x)=.x29n concluzie:x2. x29 dx=x(. x29)' dx==x .x29. x29 dx=x .x29I , Dar I =I1I2--I =x.x29I 9lnx+. x29--I =12( x.x299lnx+. x29)+Formul general:.x2a2dx=12 ( x . x2a2a2lnx+.x2a2)+, x|a , a , a>013. I =.x2+9dx ; I =?Solutie:I =x29.x2+9 dx==x2. x2+9 dx_I1+9dx. x2+9_I2I2=9ln( x+. x2+9)+Tem : Calculai I1 folosind ex12Finalizare : I =12 ( x. x2+9+9ln .x2+9)+14..9x2dx , x(3,3)Solutie:I =.9x2dx=9x2.9x2 dx==91.9x2 dx_I1x2.9x2 dx_I2I1=9arcsin(x3 )+I2=xx.9x2 dxObservmc: ( .9x2)' = x.9x2Deci I2 se poate calcula prin pri astfel :I2=x(.9x2)' dx=x.9x2+.9x2dxFinalizare :I =I1I2=9arcsin(x2 )+x .9x2I --I =12( x .9x2+9arcsin x3 )+Formul general:.a2x2dx=12 ( x .a2x2+a2arsin xa )+ x|a , a , a>015.xe2xdx , xSolutie :Notm f ( x)=x si g ' ( x)=e2x- f ' ( x)=1 si g( x)=1 2 e2xI =xe2xdx=1 2 x (e2x) ' dx==1 2 xe2x1 2 e2xdx==1 2 xe2x1 4 e2x+ --I =1 2 e2x( x1 2 )+I =1 2 e2x(2x12 )+16.x .x29dx , x>3Solutie:I =x . x29dx=x( x29).x29 dx==x3. x29 dx_I19x.x29 dx_I2unde I2=9.x29Pentru a calcula I1 notm f ( x)=x2si g ' ( x)=x. x29 -- f ' ( x)=2x si g( x)=.x29Deci :I1=x2(.x29) ' dx=x2. x292x . x29dx==x2. x292 II =I1I2=x2.x292I9.x29I =13( x29). x29+17.excos( x)dx , xSolutie :Notm f ( x)=cos( x) si g ' ( x)=ex- f ' ( x)=sin( x) si g( x)=exIntegrala devine:I =excos( x)dx=(ex) ' cos( x) dx==excos( x)ex(sin( x)) dx==excos( x)+exsin( x)dx_I'Pentru a calcula integrala I ' folosim iari formula de integrare prin pri astfel :f ( x)=sin( x) si g ' ( x)=ex- f ' ( x)=cos( x) si g ( x)=exI ' =(ex)' sin( x) dx=exsin( x)excos( x) dxn colncluzie :I =excos ( x)+exsin( x)I --I =ex2 (cos ( x)+sin( x))+18.arcsin( x) dx , x(1,1)Solutie:Alegem f ( x)=arcsin( x) si g ' ( x)=1 - f ' ( x)=1.1x2 si g( x)=xAsadar :I =arcsin( x)dx=( x)' arcsin( x)dx==xarcsin( x)x.1x2 dxObservmc: (.1x2)' =x.1x2 , n concluzie:I =xarcsin( x)+(.1x2)' dx= x codt arcsin( x)+.1x2+19.sin2( x) dx , xSolutie :Met I : Notm f ( x)=sin( x) si g' ( x)=sin( x)-- f ' ( x)=cos( x) si g ( x)=cos( x)I =sin( x)sin( x) dx=sin( x)(cos( x)) dx==sin( x) cos( x)+cos2( x) dx= Dar cos2( x)=1sin2( x) deci :cos2( x) dx=dxsin2( x) dx Finalizare :I =sin( x) cos( x)+xI , dar sin( x)cos( x)=sin(2x)2Deci :I =x214 sin(2x)+Met II : Notm f ( x)=sin2( x) si g ' ( x)=1-- f ' ( x)=2sin( x)cos( x) si g( x)=xI =( x)' sin2( x) dx=xsin2( x)2xsin( x) cos( x)dxI =xsin2( x)xsin(2x) dxFolosim iari formula de integrare prin pri:Notm f ( x)=x si g' ( x)=sin(2x) - f ' ( x)=1 si g( x)=12 cos (2x)x(12 cos(2x))' dx=12 xcos(2x)+12cos (2x) dxI =xsin (2x)dx=i I =xsin2( x)+12 xcos(2x)14 sin( 2x)+==x2 ( 2sin2( x)+cos(2x))14 sin(2x)+Dar cos (2x)=cos2( x)sin2( x) , dec:2sin2( x)+cos (2x)=2sin2( x)+cos2( x)sin2( x)=1Finalizare :I =x214 sin (2x)+20.arctg( x) dx , xSolutie:Folosimnotaia: f ( x)=arctg ( x) si g ' ( x)=1 - f ' ( x)=11+x2 si g ( x)=xObinem:I =arctg( x) dx=( x)' arctg( x) dx=xarctg( x)x1+x2 dxPrintr-o oarecare intuiie matematic observm c:|12 ln(1+x2) ' =x1+x2 , aadar :I =xarctg( x)12 ln (1+x2)+Exerciii propuseCalculai integralele:1.xexdx , x2.x2e3xdx , x3.( x1)2exdx , x 3.( x33x+2) exdx , x5.( x2)2e2xdx , x 6.xcos ( x) dx , x 7.x2cos( x) dx , x8.cos2( x) dx , x 9.e2xsin ( x) dx , x 10..x225 dx , x >511..x2+196 dx , x 12..36x2dx , x (6,6)13.x . x225 dx , x>514.ex(cos( x)) dx , x 15.arccos( x) dx , x (1,1)16.arcctg ( x) dx , xMetoda substituieiPrima metod de schimbare de varibil Probleme rezolvate:S se calculeze, folosind prima metod de schimbare de variabil, primitivele urmtoarelor funcii:1. f ( x)=2 x+1x2+x+7 , xSolutie :Notm x2+x+7=t si derivm:( x2+x+7)' dx=t ' dt - ( 2 x+1) dx=dtIntegrala devine :I =2 x+1x2+x+7 dx=dtt =lnt+Revenind la substituia fcut avem:I =ln ( x2+x+7)+2. f ( x)=2 x+3x2+3 x+1 , xSoltie :Notam x2+3 x+1=t i derivm:( x2+3 x+1) ' dx=t ' dt - (2 x+3)' dx=dtIntegrala devine :I =2x+3x2+3 x+1 dx=dtt =lnt+n final revenim la substituie :I =ln ( x2+3 x+1)+3. f ( x)=4 x+2x2+x+2 xSolutie :Notam: x2+x+2=t astfel :( x2+x+2)' =t ' dt - ( 2 x+1)' dx=dt 2 - ( 4 x+2) dx=2 dtIntegrala devine :I =2 t dt =2lnt+=ln t2+Finalizare :I =2ln ( x2+x+2)2+4. f ( x)=sin( x)1+cos2( x) xSolutie :Notam cos( x)=t , derivam:sin( x) dx=dt - sin( x) dx=dtDeci : I =sin( x)1+cos2( x) dx=dt1+t2==arctg(t )+Finalizare :I =arctg(cos( x))+5. f ( x)=tg ( x) , x(0,n2 )Solutie :Notam cos( x)=t , derivam:sin( x) dx=dt - sin( x) dx=dtObs : Am folosit faptul c tg( x)=sin( x)cos( x) astfel :I =tg ( x) dx=sin( x)cos( x) dx=dtt =ln(t )+Finalizare :I =ln(cos( x))+6. f ( x)=1+tg2( x)tg ( x) , x(0,n2 )Solutie:Met I :I =(1 tg ( x)+tg2( x)tg ( x) )dx=(1 tg ( x)+tg ( x)) dx=dxtg ( x)_I1 +tg ( x)dx_I2 I1=ctg( x)dx=cos( x)sin( x) dxNotam sin( x)=t - cos( x)dx=dt -I1=dtt =lnt+=ln(sin ( x))+I2=tg ( x)dx=sin ( x)cos( x) dxPenru a rezolva integrala I2 vom proceda n mod analogTem: Rezolvai integrala I2 Trebuie s gsii c : I2=ln(cos( x))+Finalizare :I =ln(sin( x))ln (cos ( x))+ sauI =ln(sin( x)cos( x))+=ln(tg ( x))+Met II :I =1+tg2( x)tg ( x) dx=1 tg ( x)(tg ( x))' dxObs : Am intuit foarte simplu faptul c:1+tg2( x)=cos2( x)cos2( x)+sin2( x)cos2( x)=sin2( x)+cos2( x)cos2( x) =1 cos2( x)=(tg ( x))'Aadar i prin urmare...Notamtg( x)=t - (tg ( x))' dx=dtI =lnt+Finalizare :I =ln(tg ( x))+7. f ( x)=x3 ex4, xSolutie:Notam x3ex4=t derivnd constatm:4 x3ex4=dt - x3 ex4dx=dt4 n acestecircumstane...I =x3ex4dx=1 4 dtt =1 4 lnt+the end... I =1 4 ln(ex4)+8. f ( x)=sin( x)cos2( x) , xSolutie:Folosimnotaiacos( x)=t - sin( x) dx=dtUtilizm formula de schimbare devariabil:I =sin( x) cos2( x) dx=t2dt =t3 3 +Revenimla schimbarea devariabil:I =cos3( x)3 +9. f ( x)=sin3( x)cos3( x) , xSolutie:Notam cos ( x)=t -sin( x) dx=dtI =sin3( x)cos3( x) dx=sin2( x)sin ( x)cos3( x) dx==(1cos2( x))sin( x)cos3( x) dx=(1t2)t3dt ==(t5t3) dt =t5dt t3dt ==t6 6 t4 4 +Finalizare :I =cos6( x)6 cos4( x)4 +10. f ( x)=tg ( x)+tg3( x) , x(n2 ,n2 )Solutie :Amintimdin ex6:(tg ( x))' =1 cos2( x)=sin2( x)+cos2( x)cos2( x) =cos2( x)cos2( x)+sin2( x)cos2( x)=1+tg2( x)Notamtg ( x)=t - (1+tg2( x))dx=dtI =tg ( x)+tg3( x) dx=tg ( x)(1+tg2( x)) dx==t dt =t2 2 +I =tg2( x)2 +=1 2 tg2( x)+!Obs:Pentru a beneficia de un punctaj maxim n cazul rezolvrii unui exerciiu matematic, trebuie s aducem soluia sub forma cea mai simpl.11. f ( x)=. x.1x3 , x(0; 1)Solutie :Notm x .x=t 2- ( x . x)2=x3=t2Derivm, ( x . x)' dx=dtDar ( x. x)' =.x+x2. x=3x2.x , deci :32 . xdx=dt -.xdx=23 dtintegrala I =. x.1x3 dx devineI ' =23 dt.1t2==23 arcsin(t )+Revenind la schimbare de variabil fcut obinem:I =23 arcsin ( x.x)+12. f ( x)=x1+x4 , xSolutie :Notam: x2=t - 2x dx=dt - xdx=dt2Integrala I =x1+x4 dx=122x1+x4 dx devine prin schimbare de variabila:I ' =12dt1+t2 dt =12 arctg(t )+Revenind la schimbarea factuta obtinem:I =12 arctg( x2)+13. f ( x)=e. x. x , x>0, xSolutie:Notam.x=t - 12. x dx=dt - dx.x=2dtIntegrala devine :I =e.x.x dx=2etdt =2et+Revenind la schimbarea factuta obtinem:I =2e. x+14. f ( x)=e2x.1e4x , x0, xSolutie :Notame2x=t - 2e2xdx=dt -e2x=t 2- e4x=t2- e2xdx=dt2n concluzie: I =e2x.1e4x dx=12 1.1t2 dt =12 arcsin (t )+Revenind la schimbareade variabil obtinem:I =12 arcsin (e2x)+15. f ( x)=etg( x)cos2( x) , x(n2 ,n2 )Solutie :Notamtg ( x)=t - dxcos2( x)=dtPrin schimbare devariabil:I =etg ( x)cos2( x) dx=etdt =et+Revenind la schimbarea fcut :I =etg ( x)+16. f ( x)=.1+x2, xSolutie :Incercamnotatia 1+x2=t - 2xdx=dt - x dx=dt2Tragem de aici concluzia c n acest caz metoda schimbrii de variabil nu ne prea surde.ncercm s folosim metoda integrrii prin pri....poate,poate...I =.1+x2dx=( x)' .1=x2dx=x .1+x2x21+x2 dx==x .1+x2(x2+1.x2+1 dx1.x2+1 dx) -- I =x.1+x2I +ln ( x+. x2+1)+2I =x .1+x2=ln( x+. x2+1)+Finalizare :I =12 ( x .1+x2+ln( x+.x2+1))+!!!!!Atentie la pag 30 ex 16'17. f ( x)=sin(2x)sin4( x)+3 , xSolutie :Alegem sin2( x)=t - 2sin( x)cos( x)dx=dtDar cunoastem faptul ca 2sin( x) cos ( x)=sin (2x) , deci :sin(2x) dx=dt iar sin4( x)=(sin2( x))2=t2Dup toate acestea...I =sin( 2x)sin4( x)+3 dx=dtt2+3== dtt2+( .3)2=1.3arctg t.3+Revenimasupra schimbarii facute :I =1.3 arctg(sin2( x).3 )+18. f ( x)=xtg ( x2) , x(n2 ,n2 )Solutie :Notam x2=t - 2xdx=dt - xdx=dt2I =x tg ( x2) dx=12tg (t ) dt ==12sin(t )cos (t ) dtFolosim o nou schimbare de variabil:cos(t )=a - sin(t )dt =da - sin(t ) dt =daI =12 daa =12 ln(a)+=ln (.( a))+=ln (.cos(t ))+n final I =ln(.cos( x2))+ sau I =ln(.cos( x2)cos( x2) )+19. f ( x)= 1x2+x+1 , xSolutie :Obs ca: x2+x+1=( x2+2x12 +14 )14+1==( x+12)2+34I = dx. x+2+x+1= dx( x+12)2+(.32 )2Notam x+12=t - dx=dtI = dt.t2+(.32 )2=lnt +.( x+12 )2+(.32 )2+n final:I =ln |( x+12)2+.( x+12)2+(.32 )2+ sauI =ln |( x+12)2+. x2+x+1+20. f ( x)= 1x ln( 2x) , x>1Solutie :Notam: ln (2x)=t - 22x dx=dt - dxx =dtI = dxx ln( 2x) ;I se transform prinschimbare devariabiln :I ' =dtt =lnt+Revenimla schimbarea fcut :I =ln (ln(2x)) ! Obs : Modulul a disparut pentru ca x>1Exerciii propuseCalculai primitivele urmtoarelor funcii, folosind prima metod de schimbare de variabil:1. f ( x)=3x+1x3+x+2 , x 2. f ( x)=2x+3x2+3x+6 , x3. f ( x)=6x+3x2+x+9 , x4. f ( x)=cos( x)1+sin2( x) , x5. f ( x)=ctg( x) , x(0,n2 )6. f ( x)=1tg2( x)tg( x) , x (0,n2 )7. f ( x)= xx2+5x+12 , x>e2,x8. f ( x)=1.xsin( .x) , x>0, x 9. f ( x)=x3x8+1 , x10. f ( x)=e. x.x , x>0, x11. f ( x)=x4ex5, xIntegrarea funciilor raionale simpleProbleme rezolvate:S se calculeze primitivele urmtoarelor funcii:1. f ( x)= 1x+1 , x1Solutie : 1x+1 dx=lnx+1+=ln(x1)+2. f ( x)= x( x+1)(2x+1) , x>1, xSolutie : Calculul primitivei acestei funcii presupune mai nti descompunerea ei n funcii raionale simple, adic:x( x+1)(2x+1)=Ax+1+B2x+1 Dupa ce aducem la acelasi numitor obtinem: x( x+1)(2x+1)=2 Ax+A+Bx+B( x+1)( 2x+1) , de fapt :x+0=x(2A+B)+A+B Trecem la identificarea coeficientilor: 2A+B=1A+B=0 pentru c coeficientul lui x este 1 iar coeficientul liber este 0.Rezolvnd sistemul obinem:A=1si B=1Ajungemla concluzia : x( x+1)(2x+1)=1x+112x+1 , prinurmare:f ( x)=(1x+112x+1)dx==dxx+1dx2x+1==ln ( x+1)12 ln (2x+1)+==ln (x+1.2x+1)+3. f ( x)= 1x2+2x+3 , xSolutie:Calculam radacinile polinomului f.voi folosi n loc de litera grecesca delta pe DD=b24ac=412=80 - f are radacini complexe.Datorit acestui fapt ncercm scrierea lui sub form de sum de ptrate.x2+2x+3=x2+2x+1+2=( x+1)2+(.2)2f ( x)= dx( x+1)2+( .2)2=1.2 arctg(x+1.2 )+ Propunator: prof. Gheorghita Adrian Stefane-mail: [email protected]