quiz1solns - MAT 104 Quiz 1 due on simple substitutions integration by parts and partial fractions 1(10 points e1\/x dx x3 We make the substitution t =

Thomas' Calculus: Early Transcendentals

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MAT 104 Quiz 1, due Feb 21, 2003 on simple substitutions, integration by parts andpartial fractions1.(10 points)Finde1/xx3dx.We make the substitutiont= 1/xand then use integration by parts.Ift= 1/x, thendt= (-1/x2)dxsodt=-t2du. Thuse1/xx3dx=-t3etdtt2=-tetdt=-tet+etdt=-tet+et+C=-e1/xx+e1/x+C.Here we have used integration by parts withu=tanddv=etdtsodu=dtandv=et.2.(10 points)Findπ/20cosx2-cos2xdx. (Hint: Use the identity sin2x+ cos2x= 1.)The denominator 2-cos2xis the same as 2-(1-sin2x) = 1 + sin2x. The integral becomesπ/20cosx1 + sin2xdx=sinπ/2sin 0du1 +u2from the subsitutionu= sinxanddu= cosx dx.So we get arctan(1)-arctan 0 =π4-0 =π4.3.(10 points)Find3x2arctanx3dx.Note: You may be used to calling the inverse tangent function tan-1instead of arctan. Bothnotations are standard. So arctanx3means exactly the same as tan-1(x3).Ift=x3thendt= 3x2dxand we getarctan(t)dt.Now use integration by parts withu= arctantanddv=dt. Thusdu=dt/(1 +

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