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Tekhnic-Integration-by-part-1-Find-x-sec-2-x-dx-2-Find-x-e-2x-dx-3-Find-ln-x-dx-4-Find-x-2-e-2x-dx-5-Find-e-x-cos-x-dx-




Question Number 156008 by zainaltanjung last updated on 07/Oct/21
Tekhnic  Integration by part  1) Find ∫x.sec^2 x dx  2) Find ∫x.e^(2x)  dx  3) Find ∫ln  x dx  4)  Find ∫x^2 .e^(2x)  dx  5)  Find ∫e^x  cos x dx
TekhnicIntegrationbypart1)Findx.sec2xdx2)Findx.e2xdx3)Findlnxdx4)Findx2.e2xdx5)Findexcosxdx
Commented by SANOGO last updated on 07/Oct/21
cool
cool
Commented by puissant last updated on 07/Oct/21
1)  A=∫x sec^2 x dx = xtanx−∫tanx  =xtanx−(−ln∣cosx∣)+C  ⇒ A= xtanx+ln∣cosx∣+C..
1)A=xsec2xdx=xtanxtanx=xtanx(lncosx)+CA=xtanx+lncosx+C..
Commented by puissant last updated on 07/Oct/21
2)  S=∫x e^(2x) dx = (1/2)x e^(2x) −(1/2)∫e^(2x) dx  =(1/2)x e^(2x) −(1/4)e^(2x) +C  ⇒ S= (1/2){x e^(2x) −(1/2)e^(2x) }+C
2)S=xe2xdx=12xe2x12e2xdx=12xe2x14e2x+CS=12{xe2x12e2x}+C
Commented by SANOGO last updated on 07/Oct/21
3)  ∫lnx=xlnx−x +c
3)lnx=xlnxx+c
Commented by puissant last updated on 07/Oct/21
3)  I=∫lnx dx    { ((u=lnx)),((v′=1)) :}  ⇒  { ((u′=(1/x))),((v=x)) :}  ⇒ I = xlnx−∫1dx  ⇒ I = xlnx − x + C
3)I=lnxdx{u=lnxv=1{u=1xv=xI=xlnx1dxI=xlnxx+C
Commented by puissant last updated on 07/Oct/21
4)  J=∫x^2 e^(2x) dx    { ((u=x^2 )),((v′=e^(2x) )) :}   ⇒   { ((u′=2x)),((v=(1/2)e^(2x) )) :}  ⇒ J = (x^2 /2)e^(2x) −∫xe^(2x) dx  ⇒ J=(x^2 /2)e^(2x) −(x/2)e^(2x) +(1/4)e^(2x) +C
4)J=x2e2xdx{u=x2v=e2x{u=2xv=12e2xJ=x22e2xxe2xdxJ=x22e2xx2e2x+14e2x+C
Commented by puissant last updated on 07/Oct/21
5)  D = ∫e^x cosx dx   { ((u=cosx)),((v′=e^x )) :}  ⇒  { ((u′=−sinx)),((v=e^x )) :}  ⇒ D = e^x cosx+∫e^x sinx dx   { ((u=sinx)),((v′=e^x )) :}  ⇒   { ((u′=cosx)),((v=e^x )) :}  ⇒ D = e^x cosx+e^x sinx−∫e^x cosxdx  ⇒ D = e^x (cosx+sinx)−D  ⇒ 2D=e^x (cosx+sinx)  ⇒ D=(e^x /2)(cosx+sinx)+C..
5)D=excosxdx{u=cosxv=ex{u=sinxv=exD=excosx+exsinxdx{u=sinxv=ex{u=cosxv=exD=excosx+exsinxexcosxdxD=ex(cosx+sinx)D2D=ex(cosx+sinx)D=ex2(cosx+sinx)+C..
Commented by tabata last updated on 08/Oct/21
3 ) with out by part     ∫ lnx dx = ∫ ( lnx + 1 − 1 )dx     = ∫ (( lnx + 1 ) − 1) dx     = ∫ d ( x lnx ) − ∫ dx    = x lnx − x + C    □ M
3)withoutbypartlnxdx=(lnx+11)dx=((lnx+1)1)dx=d(xlnx)dx=xlnxx+C◻M

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