2008 Test 3 solution

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10 Apr 2012
School
Department
Course
Professor
MAT137Y, 2008–2009 Winter Session, Solutions to Term Test 3
1. Evaluate the following integrals.
(10%) (i) Zln8
0
ex8+exdx.
Let u=8+ex, then du =exdx, so
Zln8
0
ex8+exdx =Z16
9
udu =2
3u3/216
9
=2
3(64 27) = 74
3.
(10%) (ii) Zdx
x2(x+1)2.
Expanding the integrand using partial fractions,
1
x2(x+1)2=A
x+B
x2+C
x+1+D
(x+1)2=Ax(x+1)2+B(x+1)2+Cx2(x+1) + Dx2
x2(x+1)2
=x3(A+C) + x2(2A+B+C+D) + x(A+2B) + B
x2(x+1)2.
Matching coefficients, we have B=1, A=2, C=2, D=1. Therefore,
Zdx
x2(x+1)2=Z2
x+1
x2+2
x+1+1
(x+1)2dx
=2ln|x|1
x+2ln|x+1|1
x+1+C=2ln
x+1
x1
x1
x+1+C.
(10%) (iii) Zsec6θdθ.
Taking out sec2θ,
Zsec6θdθ=Zsec4θsec2θdθ=Z(tan2θ+1)2sec2θdθ.
Let u=tanθ, then du =sec2θdθ; giving us,
Z(u2+1)2du =Zu4+2u2+1du =1
5u5+2
3u3+u+C
=1
5tan5θ+2
3tan3θ+tanθ+C.
(10%) (iv) Zcos(lnx)dx.
This is SHE 8.2 #38, which was assigned in Problem Set 10.
1
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