MAT 137Y 2004-2005, Solutions to Term Test 3

1. Evaluate the following integrals.

(10%) (i) Zsec2x

tanx+1dx.

Let u=tanx+1. Then du =sec2x dx. Therefore,

Zsec2x

tanx+1dx =Zdu

u=ln|u|=ln|tanx+1|+C.

(10%) (ii) Zp9−x2dx.

Let x=3sinθ. Then dx =3cosθdθ. This gives

Z3cosθ·3cosθdθ=Z9cos2θdθ=9Z1+cos2θ

2dθ=9

2θ+9

4sin2θ.

Substituting back, we see that sin2θ=2sinθcosθ=2·x

3·√9−x2

3, so

Zp9−x2dx =9

2sin−1x

3+1

2xp9−x2+C.

(10%) (iii) Z2x2−x+4

x(x2+4)dx.

We decompose the rational function using partial fractions:

2x2−x+4

x(x2+4)=A

x+Bx+C

x2+4=⇒A(x2+4) + Bx2+Cx =2x2−x+4=⇒

A+B=2

C=−1

4A=4,

so A=1, B=1, and C=−1. Thus,

Z2x2−x+4

x(x2+4)dx =Z1

x+x−1

x2+4dx =ln|x|+1

2ln(x2+4)−1

2tan−1x

2+C.

2.

(10%) (i) Find the volume generated by rotating the region bounded by the curves y=ex,y=e−x,x=1

about the y-axis.

Since the region is being rotated about the y-axis, we use shells. Since ex>e−xfor x>0, the

volume is

V=Z1

02πx(ex−e−x)dx

Integrating by parts, we let u=xand dv =ex−e−xdx. Then du =dx and v=ex+e−x. Hence

V=Z1

02πx(ex−e−x)dx =2πx(ex+e−x)

1

0−Z1

0ex+e−xdx

=2πe+e−1−hex−e−xi1

0

=2πe+e−1−e+e−1=4πe−1.

1