Calculus II
Practice Problems 4: Answers
1. Integrate
ln x 2 dx.
du 2 ln xdx x dv dx v ln x dx x ln x 2 ln xdx Answer. We integrate by parts, using u
2
ln x
2
x:
2
As we saw in example 9 (another integration by parts):
x C so
ln x dx x ln x 2 x ln x x
C
ln xdx
1
2. Integrate
x ln x
2
2
x2 lnxdx.
Answer. Let u
ln x du
x 2 dx v
dx x dv
x2 ln xdx
3. Integrate
x3
ln x
3
x3 3 :
x2
dx
3
x3
ln x
3
x3
9
C
arccos xdx.
dx 1 x dv dx v x :
x
2
arccosxdx x arccosx 1 x dx This last we integrate by the substitution w 1 x dw 2xdx :
x
1
1 x dx 2 w dw w C Answer. Let u
2
arccosx du
2
2
1 2
1 2
2
Putting this back in (2) we obtain
arccosxdx
x arccosx
1 4. If the region in the first quadrant bounded by the curves y
what is the volume of the resulting solid?
1, y
x2
C
e x and x
1 is rotated about the y-axis,
Answer. The region is drawn in figure 1. One can sweep out the volume in the y-direction, using the method
of washers, or in the x direction, using the method of shells.
3.5
3
2.5
2
1.5
1
0.5
0
0
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0.2
0.4
0.6
0.8
1
1.2
Figure 1
Washers. Here, sweeping in the y direction, the differential of volume is dV
radius is R 1, and the smaller radius is r ln y. Thus
Volume
π
e
1
1 lny dy π y y lny 2
2
e
using the solution to problem 1 above (equation (1)). Then
Volume
π e
2e
2e
2y ln y
1 2 π
1
2π
0
1 dx x ex
2π xex
ex
x2
2
π R2
2y Shells. Now, we sweep out along the x-axis, and the differential of volume is dV
and the height of the shell is e x 1. Thus
Volume
r dy.
2
The larger
e
1
2π rhdx. The radius is x,
1
0
using the formula obtained in example 5 of the Notes. We get
Volume
5. Integrate
1 e
0 1 2
2π e
π
sec3 xdx.
1 tan x:
3
sec udu tan x 1 secxdx tan x sec xdx sec xdx in example 3 of the notes, so we concentrate on the first integral. If we
The last integral was computed
write tan x sec xdx tan x sec x tan x dx, then we can integrate by parts with the substitution u tan x du sec xdx dv sec x tan xdx v sec x. Then
tan x sec xdx sec x tan x sec xdx Answer. First, we use the identity sec 2 x
3
2
2
2
2
2
2
3
It appears we’re back where we started, but not exactly. Substitute this in (3) to get:
sec3 udu
sec x tan x
sec3 xdx
sec xdx
Moving the second term to the left hand side, and dividing by 2, we have the result
sec3 udu
1
sec x tan x
2
sec xdx
1
sec x tan x
2
Incicentally, the expression found in example 3 of the Notes for
to the formula found in most integral tables:
4
sec xdx
using this sequence of identities:
ln
1
1
sinx
sinx
2 ln
ln
ln sec x
1
1
1
1
ln
4
1
tanx
C C ln 1 sin x cos x
1 cossinx
x
sinx
sinx
sec xdx is not the usual one, but is equivalent
sinx 2
sin2 x
2 ln sec x
tanx
2
Finally, this gives the answer to our problem:
5
sec3 udu
ln sec x 1
sec x tan x
2
tanx
C Answer. For part (a), we seek constants A B such that
x 1 A B x x 3
x x 3
6. Integrate (a)
x 1 dx
(b)
xx 3
x 1 dx
x2 x 3
2Bx.
3.
Put the right hand side over a common denominator and equate numerators, getting x 1 A x 3
Now substituting x 0, we get 1 3A, so A 1 3. Substitute x
3 to get 3 1 B 3 , so B
Thus
x 1 dx 1 dx 2
dx
1
2
1
ln x 3 C
ln x x 3 2 C
ln x
xx 3
3
x
3 x 3 3
3
3
For part (b), we seek constants A B C such that
x2
x 1
x 3
A
x
B
x2
x C 3 Put the right hand side over a common denominator and equate numerators, getting x 1 Ax x 3 B x
3 Cx2 . Now substituting x 0, we get 1 3B, so B 1 3. Substitute x
3 to get 3 1 C 9 , so
C
2 9. That uses up the roots, so to find A we have to equate coeffiecients. Equating the coefficients of
x2 we get 0 A C, so A 2 9. Now, we can integrate:
x 1 dx
x2 x 3
7. Integrate
dx
x 1 x 2
2 9
dx
x
2
.
1 3
dx
x2
2 9
dx
x 3
2
ln x
9
1
x
3
1
2
ln x
9
3
C 1
x A 1 x B 2 x C2 x 1 x 2 side over a common denominator
Put the right hand
and equate numerators, getting 1 A x 2 B x 1 x 2 C x 1 . Substitute x 1: 1 A 9 , so A 1 9. Substitute x 2: 1 C 3 , so C 1 3.
Now equate coefficients of x : 0 A B, so B 1 9. Thus
1 9 dx 1 9 dx dx
1 3 dx x 1 x 2 x 1
x 2
x 2
19 ln xx 12 13 x 2 C x 1 dx .
8. Integrate
x 1 x 3
Answer. We seek constants A B C such that
x 1 A Bx C x 1 x 3
x 1 x 1
x 3
Put the right hand side over a common denominator
and equate
numerators, getting x 1 A x 3 ! Bx x 3 " C x 1 . Substitute x 3 to get 3 1 C 3 1 , giving C 4 5. To find A and B we
must equate coefficients. For the constant term this gives 1 3A C, so A 3 5. Equating coefficients
of x gives us 0 A 3B, so B A 3 1 5. Thus
x 1 dx 3 dx 1 xdx 4 dx
x 1 x 3
5 x 1 5 x 1 5 x 3
35 arctanx 101 ln x 1 45 ln x 3 C Answer. We seek constants A B C such that
2
2
2
2
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
x2 dx
.
9 x
Answer. We use the substitution indicated in figure 2. Then
9. Integrate
2
x
3 sin u dx
3 cos udu
9
x2
3 cos u
3
x
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u
9
Figure 2
x2
The integral becomes
xdx
1
2
9 sin2 u 3 cos udu
9
3 cosu
9 x2
This integral requires the half-angle formula; using that we obtain:
sin2 udu
so
1
xdx
9
# 1
u
2
cos 2u du
9
u
2
1
sin 2u
2
sin2 udu
1
sin 2u
2
C C
x2
Now, to express this in terms of x, we need the double angle formula sin 2u
9
92 arcsin 3x x 99 x C 2 sin u cosu, giving finally:
2
xdx
x2
x2 dx
.
9 x
Answer. We use the substitution indicated in figure 3.
10. Integrate
2
9
x2
x
PSfrag replacements
u
3
Figure 3
Then
x
3 tanu
and the integral becomes
x2 dx
9
x2
We calculate this integral by parts: v
dx
3 sec 2 udu
9 tan2 u 3 sec2 udu
3 sec u
tan u sec u
1 sec u tanu ln sec u 2
x2
9
tan u sec u
tanu
C 3 sec u
tan2 u sec2 udu
sec 2 udu dw
tan u dv
tan2 u sec2 udu
9
sec u tan udu w
x2 dx
9 x2
9
2
9
x2 x
3
3
1
x
2
9
1
sec u tanu
2
ln sec u ln 9 x x C 3
3
x2
2
9 x C 9
ln x
2
sec u:
sec3 udu
Resubstituting back, to get an expression in terms of x:
x2 dx
9 x2
2
or
tan u
C