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MATH 16200-33: HONORS CALCULUS-2
HOMEWORK 7: SOLUTIONS
Problem 19-3. Integrate by parts.
Z
2
(ii)
x3 ex dx.
Solution. Using the substitutions
u = x2 ,
du = 2x dx,
2
dv = xex ,
v=
1 x2
e ,
2
we have
Z
3 x2
x e
2
1
dx = x2 ex −
2
Z
2
xex dx =
1 x2 2
e
x −1 .
2
Z
(viii)
cos(log x).
Solution. Let I =
R
cos(log x). Using the substitutions
u = cos(log x),
du = −
dv = dx,
sin(log x)
dx,
x
v = x,
we have
Z
I = x cos(log x) +
sin(log x) dx.
Applying the substitutions
u = sin(log x),
du =
dv = dx,
cos(log x)
,
x
v = x,
we compute
Z
Z
sin(log x) dx = x sin(log x) − cos(log x) = x sin(log x) − I.
Solving for I yields
I=
x cos(log x) + x sin(log x)
.
2
1
Z
(x)
x(log x)2 dx.
Solution. Using the substitutions
u = (log x)2 ,
du =
2 log x
dx,
x
x2
,
2
Z
Z
1 2
2
2
x(log x) dx = x (log x) − x log x dx.
2
dv = x dx,
we have
v=
Applying the substitutions
u = log x,
dv = x dx,
du =
dx
,
x
v=
x2
,
2
we compute
Z
1
1
x log x dx = x2 log x −
2
2
Z
x dx =
1 2
x2
x log x − .
2
4
Thus
Z
1
x(log x) dx = x2 (log x)2 −
2
2
=
1 2
x2
x log x −
2
4
1
1
x2
+ x2 (log x)2 − x2 log x .
4
2
2
Problem 19-4. Integrate using trigonometric substitution.
Z
dx
√
(ii)
1 + x2
Solution. Using the substitution x = tan u, dx = sec2 u du:
Z
Z
Z
dx
sec2 u du
sec2 u du
√
√
√
=
=
1 + x2
1 + tan2 u
sec2 u
Z
= sec u du
= log(sec u + tan u)
= log(sec(tan−1 x) + tan(tan−1 x))
p
= log( 1 + x2 + x) .
2
Z
(iv)
dx
x x2 − 1
√
Solution. Using the substitution x = sec u, dx = sec u tan u du:
Z
Z
Z
Z
sec u tan u du
sec u tan u du
dx
√
√
√
=
=
= du = sec−1 x .
x x2 − 1
sec u sec2 u − 1
sec u tan2 u
(viii)
Z p
1 − x2 dx
Solution. Using the substitution x = sin u, dx = cos du:
Z p
Z p
Z
2
2
1 − x dx =
1 − sin u cos u du = cos2 u du
Z
1 + cos 2u
1
sin 2u
=
du
=
u+
2
2
2
−1
sin(2 sin x)
1
sin−1 x +
=
2
2
2 sin(sin−1 x) cos(sin−1 x)
1
−1
sin x +
=
2
2
p
1
=
sin−1 x + x 1 − x2 .
2
Problem 19-5. Integrate.
Z
dx
(ii)
1 + ex
Solution. Using the substitution u = 1 + ex , du = ex dx:
Z
Z
Z
dx
(1 + ex ) − ex
ex
=
dx
=
1
−
dx
1 + ex
1 + ex
1 + ex
Z
Z
ex
du
=x−
=x−
1 + ex
u
= x − ln(u)
= x − ln(1 + ex ) .
Z
(iv)
√
dx
1 + ex
3
Solution. Using the substitution u =
Z
dx
√
=
1 + ex
Z
2 du
=
ex
Z
2 du
=
u2 − 1
√
1 + ex , du =
Z x
√e
2 1+ex
1
1
−
u−1 u+1
dx,
√
u − 1
1 + ex − 1 .
= ln √
du = ln u + 1
1 + ex + 1 Problem 19-26. Let φ be a nonnegative integrable function such that φ(x) = 0
R1
for |x| > 1 and such that −1 φ = 1. For h > 0, let
φh (x) =
1
φ(x/h).
h
(a) Show that φh (x) = 0 for |x| ≥ h and that
Rh
φh = 1.
−h
Proof. If |x| ≥ h then |x/h| ≥ 1, so φ(x/h) = 0, and
φh (x) =
1
φ(x/h) = 0.
h
Using the substitution u = x/h, du = dx/h:
Z
h
φh (x) dx =
−h
1
h
Z
h
Z
1
φ(x/h) dx =
−h
φ(u) du = 1.
−1
(b) Let f be an integrable on [−1, 1] and continuous at 0. Show that
Z
lim+
h→0
1
−1
Z
φh f = lim+
h→0
h
φh f = f (0).
−h
Proof. Part (a) shows that the first two integrals are equal. Now let ε > 0 be
given, and choose δ > 0 so that |f (x) − f (0)| < ε whenever |x| < δ. Then for
0 < h < δ,
Z
Z
Z h
Z h
h
h
φh f = φh f (0) −
φh f = φh (f (x) − f (0)) dx
f (0) −
−h
−h
−h
−h
Z h
≤
φh |f (x) − f (0)| dx
−h
Z h
≤ε
φh dx = ε
−h
4
Problem 19-42. Find the value of
R∞
2
e−x dx.
0
(a) Show that
1
Z
Z
0
2 4
2n
· · ··· ·
,
3 5
2n + 1
(1 − x2 )n dx =
0
∞
π 1 3
1
2n − 3
dx = · · · · · · ·
.
(1 + x2 )n
2 2 4
2n − 2
Proof. Using the substitution x = cos u, dx = − sin u du:
Z
1
Z
2 n
0
2n
(1 − x ) dx =
0
(sin
Z
π/2
=
π/2
u)(− sin u) du =
sin2n+1 u du
0
2 4
2n
· ···
.
3 5
2n + 1
Using the substitution x = cot u, dx = − csc2 u du:
Z ∞
Z 0
1
−1
2n
dx =
(sin u)
du
(1 + x2 )n
sin2 u
0
π/2
Z π/2
=
sin2(n−1) u du
0
2n − 3
π 1 3
.
= · · ···
2 2 4
2n − 2
(b) Prove, using the derivative, that
2
1 − x2 ≤ e−x
2
1
e−x ≤
1 + x2
for 0 ≤ x ≤ 1,
for 0 ≤ x.
2
Proof. Let f (x) = e−x + x2 − 1. Then f (0) = 0 and
2
2
f 0 (x) = 2xe−x ex − 1 ≥ 0,
2
so f (x) ≥ 0 for all 0. In particular, 1 − x2 ≤ e−x for 0 ≤ x ≤ 1. For the second
inequality, note that 1 + y ≤ ey for all y ≥ 0. Setting y := x2 ,
2
2
e−x ≤
1 + x2 ≤ ex ,
for x ≥ 0.
5
1
1 + x2
(c)
Solution. We have
Z 1
Z
(1 − x2 )n dx ≤
0
1
2
e−nx dx ≤
Z
0
∞
2
e−nx dx ≤
0
Z
∞
0
1
dx,
(1 + x2 )n
so that
Z 1
Z ∞
2
2 4
2n
π 1 3
2n − 3
−nx2
· ···
≤
e
dx ≤
e−nx dx ≤ · · · · ·
.
3 5
2n + 1
2
2
4
2n − 2
0
0
√
√
Setting y = nx, dy = n dx:
√
2 4
2n
1
· ···
≤√
3 5
2n + 1
n
Z
n
0
2
1
e−y dy ≤ √
n
Z
∞
2
e−y dy ≤ n
0
π 1 3
2n − 3
· · ···
.
2 2 4
2n − 2
Thus
√ 2 4
2n
n · ···
≤
3 5
2n + 1
√
Z
n
2
e−y dy ≤
Z
0
∞
2
e−y dy ≤
0
√ π 1 3
2n − 3
n · · ···
.
2 2 4
2n − 2
(d) Now use Problem 41(d) to show that
√
Z ∞
π
−y 2
.
e
dy =
2
0
Proof.
Problem 15-2(vi). Find
lim
x→0
1
1
−
x sin x
using l’Hôpital’s Rule.
Solution. Note that
1
1
sin x − x
−
=
.
x sin x
x sin x
Evaluating the expression at 0 gives the indeterminate 0/0, so we may apply
l’Hôpital’s Rule.
1
1
sin x − x
cos x − 1
lim
−
= lim
= lim
.
x→0 x
x→0
x→0 x cos x + sin x
sin x
x sin x
Evaluation at 0 again gives the indeterminate 0/0, so
cos x − 1
− sin x
lim
= lim
= 0.
x→0 2 cos x − x sin x
x→0 x cos x + sin x
6
Problem 15-33.
(a) Show that
1
1
x
sin(k + )x − sin(k − )x = 2 sin cos kx.
2
2
2
Proof. Applying the double angle formulas,
1
1
sin(k + )x − sin(k − )x
2
2
x x
x
x
= sin kx cos + cos kx sin
− sin kx cos − cos kx sin
2
2
2
2
x
= 2 sin cos kx
2
(b) Conclude that
1
sin(n + 1/2)x
+ cos x + cos 2x + · · · + cos nx =
.
2
2 sin x2
Proof. From part (a),
n 1
1
x X
1
sin(n + )x − sin =
sin(k + )x − sin(k − )x
2
2
2
2
=
k=1
n
X
2 sin
k=1
x
cos kx
2
n
= 2 sin
xX
cos kx.
2
k=1
Rearranging gives
n
X
sin(n + 1/2)x − sin x2
=
cos kx,
x
2 sin 2
k=1
sin(n + 1/2)x
1
= +
2 sin x2
2
as desired.
7
n
X
k=1
cos kx,
Problem 19-43.
(b) Use Problem 15-33 to show that
Z π
sin(n + 12 )t
dt = π.
sin 2t
0
Proof. We compute,
Z
0
π
sin(n + 12 )t
dt =
sin 2t
Z
π
1+
0
n
X
!
2 cos kt
k=1
dt = π+
n
X
2(sin(πk) − sin(0k))
k=1
k
= π.
(c) Prove that
Z
lim
λ→∞
0
π
1
2
1
sin(λ + )t
−
dt = 0.
2
t
sin 2t
Proof. From Problem 15-2, the expression inside the parentheses can be extended to a continuous function on [0, π]. Apply the Riemann-Lebesgue lemma.
(d) Use the substitution u = (λ + 12 )t and use part (b) to show that
Z ∞
sin x
π
= .
x
2
0
Proof. From part (c),
Z π
Z π
sin(λ + 12 )t
2 sin(λ + 12 )t
dt
= 0,
dt −
lim
λ→∞
t
sin 2t
0
0
and by part (b), the limit of the latter integral as λ → ∞ exists and equals π.
Hence
Z π
Z π
2 sin(λ + 12 )t
sin(λ + 21 )t
lim
dt = lim
dt = π.
λ→∞ 0
λ→∞ 0
t
sin 2t
Using the substitution u = (λ + 21 )t, du = λ + 1/2 dt on the former,
Z
π = lim
λ→∞
0
π
Z (λ+1/2)π
(λ + 21 )
2 sin(λ + 12 )t
du
dt = lim
2 sin u ·
·
λ→∞ 0
t
u
λ + 12
Z ∞
sin u
=2
du.
u
0
8