Cal 2 test 1 solutions - Math 1592 Calculus II Sample Test 1 Solutions tan-1 x dx 1 Integration by parts If u = tan-1 x v = x dx dv = dx du = 1 x2 then

Cal 2 test 1 solutions - Math 1592 Calculus II Sample Test...

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Math 1592 Calculus II-Sample Test 1 Solutions1.Ztan-1x dxIntegration by parts. Ifu=tan-1x,v=x,du=dx1+x2,dv=dx,thenZtan-1x dx=xtan-1x-Zx1+x2dxand ifu=1+x2, thendu=2xdxsoZtan-1x dx=xtan-1x-12Zduudx,=xtan-1x-12ln|u|+c,=xtan-1x-12ln|1+x2|+c.2.Zlnxx2dxIntegration by parts. Ifu=lnx,v=-1x,du=dxx,dv=dxx2,thenZlnxx2dx=-lnxx+Zdxx2=-lnxx-1x+c.3.Z10xexdxIntegration by parts. Ifu=x,v=ex,du=dx,dv=exdx,thenZ10xexdx=xex|10-Z10exdx=xex|10-ex|10= (e-0)-(e-1) =1.1
4.Zsin-1x dxIntegration by parts. Ifu=sin-1x,v=x,du=dx1-x2,dv=dx,thenZsin-1x dx=xsin-1x-Zx1-x2dxand ifu=1-x2, thendu=-2xdxsoZsin-1x dx=xsin-1x+12Zduudx,=xsin-1x+u+c,=xsin-1x+p1-x2+c.5.Z10dx1-x2This integral is improperlimb1Zb0dx1-x2=limb1sin-1xflflflb0=limb1sin-1b=π2.6.Z51lnx dxIntegration by parts. Ifu=lnx,v=x,du=dxx,dv=dx,thenZ51lnx dx=xlnx|51-Z51dx=xlnx|51-x|51=5 ln 5-4.7.Zsec3x dxIntegration by parts. Ifu=secx,v=tanx,du=secxtanx dx,dv=sec2x dx,thenZsec3x dx=secxtanx-Zsecxtan2x dx=secxtanx-Zsecx(sec2x-1)dx=secxtanx-Zsec3x dx+Zsecx dx2
so2Zsec3x dx=secxtanx+ln|secx+tanx|,soZsec3x dx=12secxtanx+12ln|secx+tanx|+c.8.Z3-3dx4-x2dxThis integral is improper because the integrand is undefined atx=-2 andx=2. So weconsiderZ3-3dx4-x2dx=Z-2-3dx4-x2+Z2-2dx4-x2+Z32dx4-x2and will show that one of the integrals (the last one) diverges.Z32dx4-x2=lima2Z3adx4-x2dx=14lima2Z3a12+x+12-xdx=14lima2(ln|2+x| -ln|2-x|)|3a=14lima2(ln 5-ln|2+a|+ln|2-a|) =-.Since one integral diverges, the entire integral diverges.9.Zsin2xcos3x dxTrig. integral. If we re-write the integral asZsin2xcos2xcosx dx,then ifu=sinx, thendu=cosx dxand we haveZsin2xcos2xcosx dx=Zu2(1-u2)du=u33-u55+c=sin3x3-sin5x5+c.

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