MAT137Y, 2008–2009 Winter Session, Solutions to Term Test 3

1. Evaluate the following integrals.

(10%) (i) Zln8

0

ex√8+exdx.

Let u=8+ex, then du =exdx, so

Zln8

0

ex√8+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,

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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=Z−2

x+1

x2+2

x+1+1

(x+1)2dx

=−2ln|x|− 1

x+2ln|x+1|− 1

x+1+C=2ln

x+1

x−1

x−1

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.

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