MAC 2312-15931–Calculus II

MAC 2312-15931–Calculus II
Spring 2003
Homework #3–Solutions
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1. Compute the following integrals. In all cases show work.
Z
(a)
x3 e−4x dx.
Solution. Integrating by parts,
Z
Z
1
3
x3 e−4x dx = − x3 e−4x +
x2 e−4x dx
4
4
µ
¶
Z
1
3
1
1
= − x3 e−4x +
− x2 e−4x +
xe−4x dx
4
4
4
2
µ
¶
Z
1 3 −4x
3 2 −4x 3
1
1
= − x e
− x e
+
− xe−4x +
e−4x dx
4
16
8
4
4
µ
¶
1 3 −4x
3 2 −4x
3 −4x
3
1
= − x e
− x e
− xe
+
− e−4x + C
4
16
32
32
4
1 3 −4x
3 2 −4x
3 −4x
3 −4x
= − x e
− x e
− xe
−
e
+ C.
4
16
32
128
Z
(b)
e2x sin(3x) dx.
Solution. This is one of these integrals where integration by parts
gives an equation for the integral, so we begin setting
Z
I = e2x sin(3x) dx.
Now,
Z
1 2x
3
(1) I =
e sin(3x) −
e2x cos(3x) dx
2
2
µ
¶
Z
1 2x
3 1 2x
3
2x
=
e sin(3x) −
e cos(3x) +
e sin(3x) dx
2
2 2
2
1 2x
3
9
=
e sin(3x) − e2x cos(3x) − I.
2
4
4
Solving for I,
Z
2 2x
3
I = e2x sin(3x) dx =
e sin(3x) − e2x cos(3x) + C.
13
13
Z
(c)
xe2x sin(3x) dx.
Solution. This integral is a minor patience test. Perhaps after some
false starts, or immediately, one should see that no substitution works
well, but that integration by
R parts has a chance. If one uses u = x,
dv = e2x sin(3x) dx, then v du might actually be doable, because
the x has disappeared. For this one has to first have the integral of
e2x sin(3x), which we found in the previous exercise. We refer to this
exercise, and use the result to find the integral we want. Integrating
by parts, with u = x, dv = e2x sin(3x) dx,
µ
¶
Z
2 2x
3 2x
2x
xe sin(3x) dx = x
e sin(3x) − e cos(3x)
13
13
µ
¶
Z
2 2x
3 2x
−
e sin(3x) − e cos(3x) dx.
13
13
This seems still horrible, but we are almost done. No more integrations are needed, they have all been done. The last integral
divides into
Z
Z
2
3
e2x sin(3x) dx −
e2x cos(3x) dx.
13
13
The first integral is the one we called I in the previous exercise and
calculated before. The second one is similar to I but, if we were very
neat and careful, it has actually been secretly calculated. Return to
the equation labelled (1) (in the previous exercise) and notice that it
can be solved to get the second integral. We get
Z
1
2
e2x cos(3x) dx = e2x sin(3x) − I.
3
3
Using the value of I found before,
µ
¶
Z
2 2 2x
3 2x
1 2x
2x
e sin(3x) −
e sin(3x) − e cos(3x) + C
e cos(3x) dx =
3
3 13
13
3 2x
2
=
e sin(3x) + e2x cos(3x) + C.
13
13
2
Now it is just a matter of putting it all together to get
µ
¶
Z
2 2x
3
xe2x sin(3x) dx = x
e sin(3x) − e2x cos(3x)
13
13
Z
Z
2
3
−
e2x sin(3x) dx +
e2x cos(3x) dx
13
13
µ
¶
µ
¶
3 2x
2
2 2x
3 2x
2 2x
e sin(3x) − e cos(3x) −
e sin(3x) − e cos(3x)
=x
13
13
13 13
13
µ
¶
2
3
3 2x
e sin(3x) + e2x cos(3x) + C.
+
13 13
13
We are done, but this looks too horrible to leave as it is in a homework (in an exam, one actually should leave it as it is, except if one
has finished everything else and there is time left). Some factoring,
collecting of terms, gives
Z
26x + 5 2x
12 − 39x 2x
xe2x sin(3x) dx =
e sin 3x +
e cos 3x + C.
169
169
Z
(d)
x2 arctan x dx.
If in this exercise you happen to run into an integral involving something like x3 over x2 +1, you might consider long division. You should
get
x3
something
= polynomial +
.
2
x +1
x2 + 1
Solution. Integrating by parts,
Z
Z
1 3
1
x3
x2 arctan x dx =
x arctan x −
dx
2
3
3
x +1
µ
¶
Z
1 3
1
x
=
x arctan x −
x− 2
dx
3
3
x +1
1 3
1
1
=
x arctan x − x2 + ln(x2 + 1) + C
3
6
6
Z
2
(e)
x3 e−2x dx.
Solution. This integral is a bit harder in a way. It can be integrated
2
by parts, using u = x2 , dv = xe−2x . Then
Z
2
2
1
v = xe−2x dx = − e−2x
4
and integration by parts gives:
Z
Z
2
1 2 −2x2 1
3 −2x2
x e
dx = − x e
+
xe−2x dx
4
2
2
2
1
1
= − x2 e−2x − e−2x + C.
4
8
3
Z
(f)
x5x dx.
Solution. If one uses that 5x = e(ln 5)x , evaluating the integral
becomes an easy integration by parts:
Z
Z
Z
1
1
x
(ln 5)x
(ln 5)x
x5 dx =
xe
dx =
xe
−
e(ln 5)x dx
ln 5
ln 5
1
1
xe(ln 5)x −
=
e(ln 5)x + C
ln 5
(ln 5)2
1
1
x5x −
5x + C.
=
ln 5
(ln 5)2
2. Consider the integral
Z
xa ln x dx.
It can be evaluated integrating by parts. Show that it can also be evaluated
by the substitution x = eu .
Solution The substitution x = eu gives dx = eu du, ln x = u; the integral
becomes
Z
Z
xa ln x dx = ue(a+1)u du.
Now one can use integration by parts. I won’t grade this problem, it
was a somewhat silly idea to have it.
3. Textbook, Section 7.2, # 20 (p. 482).
Solution There are many tricks possible, but a good one is to multiply
numerator and denominator of the integrand by cos x + 1:
cos x + 1
cos x + 1
1
=
=−
cos x − 1
cos2 x − 1
sin2 x
so that
Z
Z
Z
dx
cos x
1
=−
dx−
csc2 x dx =
+cot x+C = csc x+cot x+C.
2
cos x − 1
sin x
sin x
.
4. Textbook, Section 7.2, # 58 (p. 483).
Solution By the method of slices,
Z
π/4
V =π
tan4 x dx.
0
To compute the integral we use the identity tan2 x = sec2 x − 1, thus
tan4 x = tan2 x(sec2 x−1) = tan2 x sec2 x−tan2 x = tan2 x sec2 x−sec2 x+1.
4
Now
Z
Z
1
tan4 x dx = (tan2 x sec2 x − sec2 x + 1) dx = tan3 x − tan x + x + C.
3
Thus
µ
V =π
¶
1
π2
2π
π/4
tan3 x − tan x + x |0 =
−
.
3
4
3
5. Textbook, Section 7.3, # 6 (p. 488).
√
Solution. We try x = 2 tan u. Then x2 + 4 = 2 sec u, dx = 2 sec2 u du.
Moreover, u = arctan(x/2), so that x = 0 implies u = 0, and x = 2 implies
u = π/4. Thus
Z 2 p
Z π/4
x3 x2 + 4 dx = 32
tan3 u sec3 u du.
0
0
We now use the strategy of page 480 of the textbook, writing
tan3 u sec3 u = (sec2 u − 1) sec2 u tan sec u
and then substitute v = sec u, dv =√sec u tan u. Noticing that u = 0
corresponds to v = 1, u = π/4 to v = 2,
Z
2
x3
p
Z
x2 + 4 dx
=
√
2
32
1
0
=
64(1 +
15
µ
(v 4 − v 2 )dv = 32
√
2)
1 5 1 3
v − v
5
3
√
¶¯¯ 2
¯
¯
¯
1
.
6. Textbook, Section 7.3, # 10 (p. 488).
Solution. Substituting x = a tan u we get
Z √ 2
Z
x − a2
1
tan2 u
dx
=
du.
x4
a2
sec3 u
One could use the strategy of the previous exercise, but it is easier to
notice that
tan2 u
= sin2 u cos u
sec3 u
so that
Z √ 2
Z
x − a2
1
1
dx = 2
sin2 u cos u du = 2 sin3 u + C.
x4
a
3a
√
Finally, if secu = x/a, then sin u = x2 − a2 /x so that
Z √ 2
(x2 − a2 )3/2
x − a2
dx
=
+ C.
x4
3a2 x3
5
7. Textbook, Section 7.3, # 20 (p. 489).
Solution. We could try x = 3 sin u, but the easiest substitution is u =
9 − x2 , with which
¯0
Z 3 p
Z
¯
1 0 1/2
1
x 9 − x2 dx = −
u du = − u3/2 ¯¯ = 9.
2 9
3
0
9
8. Textbook, Section 7.3, # 30 (p. 489).
Solution. We begin substituting x = et ; then e2t = x2 , t = ln x and
dt = (1/x) dx. Thus
Z p
Z √ 2
x −9
2t
e − 9 dt =
dx.
x
Now we substitute x = 3 sec u to
Z p
Z √ 2
Z
x −9
2t
e − 9 dt =
dx = 3 tan2 u du
x
Z
= 3 (sec2 u − 1) du = 3 tan u − 3u + C.
Now
u = sec−1 (x/3) and one sees that if sec u = x/3, then tan u =
√
2
x − 9/3; using x = et we get our final answer
Z p
p
et
e2t − 9 dt = e2t − 9 − 3 sec−1 + C.
3
The solution can also be written in the form
√
Z p
p
e2t − 9
−1
2t
2t
e − 9 dt = e − 9 − 3 sin
+C
et
and in several other ways.
9. Textbook, Section 7.3, # 34 (p. 489).
Solution. A picture should definitely√be drawn. The hyperbola intersects the line x = 3 at the points (3, ±3 5/2) and the line y = 0 at (2, 0).
The region can be described from the point of view of the x-axis or the
y-axis either by
(a)
R = {(x, y) : 2 ≤ x ≤ 3, −
3p 2
3p 2
x −4≤y ≤
x − 4},
2
2
or by
(b)
√
√ p
3 5
3 5
36 + 4y 2
R = {(x, y) : −
≤y≤
,
≤ x ≤ 3}.
2
2
3
6
The first point of view is the simpler one; using it the area works out to
¶
Z 3µ p
Z 3p
3
3p 2
A=
x2 − 4 − (−
x − 4) dx = 3
x2 − 4 dx.
2
2
2
2
√
Substituting x = 2 sec t, then dx = 2 tan t sec t, x2 − 4 = 2 tan t. To see
how the limits change, when x = 2 then t should be such that 2 sec t = 2
or sec t = 1, giving t = 0. When x = 3 we get sec t = 3/2 or cos t = 2/3,
we’ll just write arccos 2/3 for t. Thus
Z
arccos 2/3
A = 12
Z
0
Now
Z
sec3 t dt
arccos 2/3
tan2 t sec t dt = 12
(sec3 t − sec t) dt.
0
Z
Z
sec t sec2 t dt = sec t tan t − tan2 t sec t dt
Z
Z
Z
= sec t tan t − (sec2 t − 1) sec t dt = sec t tan t − sec3 t dt + sec t dt,
=
thus
Z
1
1
sec t dt = (sec t tan t) +
2
2
R
Thus, using the known form of sec t dt
3
Z
sec t dt.
¯arccos 2/3
Z arccos 2/3
¯
¯
A = 12
(sec t − sec t) dt = 6 sec t tan t¯
−6
sec t dt
¯
0
0
0
¯arccos 2/3
¯
= (6 tan t sec t − 6 ln(sec t + tan t))¯¯
0
√
√
3+ 5
3 5
− 6 ln(
),
=
2
2
√
where we used that tan(arccos(2/3)) = 5/2, sec(arccos 2/3) = 3/2. Thus
√
√
3 5
3+ 5
A=
− 6 ln(
).
2
2
Z
arccos 2/3
3
On the other hand, by the second description, the area works out to
!
!
p
p
Z 3√2 5 Ã
Z 3√2 5 Ã
36 + 4y 2
2 9 + y2
A=
3−
dy = 2
3−
dy.
√
3
3
−325
0
Thus
√
Z
3
A=9 5−4
0
7
√
5
2
p
9 + y2
dy.
3
This time, instead
of doing a trigonometric substitution, we’ll do y =
p
3 sinh u; then 9 + y 2 = 3 cosh u, dy = 3 cosh u du. The limits are a bit
trickier. √
To get y = 0, we take u =
√ 0. But the value of u that corresponds
to y = 3 5/2 satisfies sinh u = 3 5/2, hence
√
eu − e−u
3 5
=
;
2
2
√
setting z = eu , thus e−u = 1/z, we
z 2 − 5z −
√get (after some arithmetic)
u
1 = 0, which has
√ the solutions ( 5 ± 3)/2.
√ Since z = e > 0, we must
u
have z = e = ( 5 + 3)/2, hence u = ln[( 5 + 3)/2].
A
=
√
Z
√
Z
√
ln[(3+ 5)/2]
9 5 − 12
=
0
√
ln[(3+ 5)/2]
9 5 − 12
√
0
√
ln[(3+ 5)/2]
Z
9 5−3
=
cosh2 u du
µ
eu + e−2u
2
¶2
du
(e2u + 2 + e−2u ) du
0
To compute this last integral, we notice that if u = ln[(3 +
√
5)/2], then
√
√ !2
7+3 5
3+ 5
=
,
2
2
√
√ !2
µ
¶2 Ã
7−3 5
2
3− 5
√
=
=
;
2
2
3+ 5
Ã
2u
=
e−2u
=
e
using this one sees without too much trouble that
Ã
√
√ !
Z ln[(3+√5)/2]
3 5
3+ 5
2u
−2u
,
(e + 2 + e
) du =
+ 2 ln
2
2
0
thus the area works out again to
Ã
√
√ !
9 5
3+ 5
A=
− 6 ln
.
2
2
10. Textbook, Section 7.3, # 35 (p. 489).
Solution. Following the hint, and referring to the picture in the text, we
see first that the area of the triangle P OQ is (r2 sin θ cos θ)/2. We used, of
course, that OQ = r cos θ, P Q = r sin θ. Because Q is situated at r cos θ
on the x-axis, the area of the region P QR is given by
Z r
p
r2 − x2 dx.
r cos θ
8
√
In this integral we substitute x = r cos t (for a change), so that r2 − x2 =
r sin t, dx = −r sin t dt, and the limits change as follows. When x = r cos θ,
then we should have r cos t = r cos θ, so t = θ; when x = r we need
r cos t = r, so t = 0. Thus
Z
r
p
Z
r2 − x2 dx
0
= −r2
r cos θ
Z
sin2 t dt = r2
θ
2
=
r
2
Z
θ
0
θ
sin2 t dt
0
r2
(1 − cos 2t) dt =
2
µ
1
t − sin 2t
2
¶¯¯θ
¯
¯
¯
0
2
=
r
1
r2
r2
(θ − sin 2θ) = θ −
sin θ cos θ.
2
2
2
2
It follows that the area equals
r2
AreaP OQ + areaP QR =
sin θ cos θ +
2
9
µ
r2
r2
θ−
sin θ cos θ
2
2
¶
=
r2
θ.
2