
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