AM1
Mathematical Analysis 1
Oct. 2011 – Feb. 2012
Exercises Lecture 3
Date: October 21
Exercise 3.1. If h 6= 0, prove the following identities hold for all x:
sin(x + h) − sin x
sin γ
=
cos(x + γ)
h
γ
cos(x + h) − cos x
sin γ
=−
sin(x + γ)
h
γ
(3.1)
(3.2)
where γ = h2 .
Solution:
We are going to show (3.1). The proof of (3.2) is left as an homework.
Recall the sum-to-product identity
sin θ − sin ϕ = 2 sin
θ−ϕ
2
cos
θ+ϕ
2
(3.3)
If we use it with θ = x + h and ϕ = x, we obtain
sin(x + h) − sin x
2
= sin
h
h
h
2x + h
cos
2
2
which is just equivalent to (3.1).
Exercise 3.2. Prove or disprove each of the following statements.
1. For all x 6= 0, we have sin 2x 6= 2 sin x.
2. For every x, there is a y such that cos(x + y) = cos x + cos y.
3. There is an x such that sin(x + y) = sin x + sin y for all y.
Ry
4. There is a y 6= 0 such that 0 sin x dx = sin y
Solution:
1. This statement is clearly false. Take indeed x = 2π, then sin 4π = 0 = 2 sin 2π.
3-1
(3.4)
3-2
2. This one is false as well. Take x = 0, then you have
cos(x + y) = cos(y) 6= 1 + cos(y) = cos(x) + cos(y)
(3.5)
3. This statement is true, and we prove it exhibiting the right x, that is 0. With this choice,
sin(x + y) = sin y = 0 + sin y = sin x + sin y
(3.6)
note that (3.6) is valid for any y.
4. This last statement is true. We prove it calculating the integral,
Z y
sin x dx = 1 − cos(y)
(3.7)
0
Then we want to exhibit a y such that
sin y = 1 − cos y
(3.8)
This equality clearly holds for y = π2 , since sin π2 = 1 and cos π2 = 0. As a matter of fact,
equation (3.8) holds for any y = π2 + 2kπ, k ∈ Z.
Rb
Exercise 3.3. Calculate the integral a sin x dx for each if the following values of a and b. In each
case interpret your result in terms of areas.
1. a = 0, b =
π
6
and a = 0, b =
π
4
2. a = 0, b =
π
3
and a = 0, b =
π
2
3. a = 0, b = π and a = 0, b = 2π
4. a = 1, b = −1 and a = − π6 , b =
π
4
Solution:
We will solve only case no.1 since the others are completely analogous and left as an easy homework.
Z π
π 6
sin x dx = 1 − cos
(3.9)
6
0
√
3
=1−
(3.10)
2
Z
0
π
4
π sin x dx = 1 − cos
4
√
2
=1−
2
(3.11)
(3.12)
3-3
Exercise 3.4. Evaluate the following integrals
1.
Rπ
2.
R
π
2
3.
R
π
2
4.
R
π
2
8.
R x2
9.
R
π
2
10.
R
π
3
0
0
0
(x + sin x) dx
(x2 + cos x) dx
(sin x − cos x) dx
| sin x − cos x| dx
Rπ
5. 0 12 + cos t dt
Rπ
6. 0 12 + cos t dt
Rx 7. −π 12 + cos t dt, with 0 ≤ x ≤ π
0
x
(t2 + sin t) dt
sin 2x dx
0
cos x2 dx
0
Solution:
π2
2
1.
Rπ
2.
R
π
2
3.
R
π
2
4.
R
π
2
5.
Rπ
6.
R 2π
R π 1
3
0 2 + cos t = 0
0
0
0
(x + sin x) =
+ 1 − cos π = 2 +
π3
24
(x2 + cos x) =
π2
2
+1
(sin x − cos x) = 1 − 1 = 0
0
| sin x − cos x| =
1
2
0
+ cos t =
R
π
4
R
π
2
π
4
√
(sin x − cos x) = 2( 2 − 1)
0
(cos x − sin x) +
1
2
Rπ
+ cos t + 2π − 21 − cos t =
π
2
3
7. We have to distinguish two cases: (a) 0 ≤ x ≤
(a)
(b)
2π
3
17π
3
+
and (b)
√
3
2π
3
< x ≤ π.
√
R 2π
R
R x 1
+ cos t = − 3 − 1 − cos t + x 2π
−π
2
−π 2
−
1
2
+ cos t =
R − 2π
R
R x 1
3
− 21 − cos t +
−π 2 + cos t = −π
1
2
Rx
+ cos t + 2π − 12 − cos t
3
2π
3
− 2π
3
17π
3
+
3+
3
the computation of (b) is left as an easy homework.
8.
R x2
x
(t2 + sin t) =
(x2 )3 −(x)3
3
+ cos(x) − cos(x3 ) =
x6 −x3
3
+ cos(x) − cos(x2 )
x
2
+ sin x
3-4
9.
R
π
2
R πϕ
R π2
x
1
sin
2x
=
sin
sin
x
=
1
here
you
have
to
take
ϕ
=
and
note
that
0
0
2
ϕ
0ϕ
0
Rπ
which in turn is equal to ϕ 0 sin x by formula (2.26) from last notes.
10.
R
π
3
0
sin 2x =
1
2
cos x2 = 2
Rπ
R
π
6
0
cos x = 1 again by means of (2.26).
Exercise 3.5. Use the identity sin 3t = 3 sin t − 4 sin3 t to deduce the integration formula
Z
x
sin3 t dt =
0
2 1
− (2 + sin2 x) cos x
3 3
Solution:
By linearity with respect to the integrand we have
Z
x
3
sin t =
0
=
=
=
=
=
=
Z x
Z x
1
3
sin t −
sin 3t
4
0
0
Z x
Z
1
1 3x
3
sin t −
sin t
4
3 0
0
Z
Z
1 8 x
1 3x
sin t −
sin t
4 3 0
3 x
1 8
1
(1 − cos x) − (cos x − cos 3x)
4 3
3
1 8
2
(1 − cos x) − sin 2x sin x
4 3
3
1 8
4
(1 − cos x) − sin2 x cos x
4 3
3
1 8 4
2
− cos x(2 + sin x)
4 3 3
which is our goal.
Exercise 3.6. Derive the identity cos 3t = 4 cos3 t − 3 cos t and use it to prove that
Z
0
Solution:
x
1
cos3 t dt = (2 + cos2 x) sin x
3
3-5
We begin verifying the required identity. By means of addition formulas
cos 3t = cos t cos 2t − sin t sin 2t
= cos t(cos2 t − sin2 t) − 2 sin2 t cos t
= cos3 t − 3 sin2 t cos t
= cos3 t − 3(1 − cos2 t) cos t
= 4 cos3 t − 3 cos t
Now we can use this formula together with the one for sin 3t from previous exercise. Thus
Z x
Z x
(4 cos3 t − 3 cos t)
(3.13)
cos 3t =
0
0
or equivalently, by linearity of the integral
Z x
Z
1 x
cos3 t =
(cos 3t + 3 cos t)
4 0
0
Z
1 1 3x
=
cos t + 3 − 3 sin x
4 3 0
1 1
sin 3x + 3 sin x
=
4 3
1
4
3
=
4 sin x − sin x
4
3
1
2
= sin x 1 − sin x
3
2 1
2
=
+ cos x sin x
3 3
(3.13a)
which is our goal.
Exercise 3.7. Prove that if a function f is periodic with period T > 0 and integrable on [0, T ],
prove that
Z T
Z a+T
f (x) dx =
f (x) dx
∀a ∈ R
(3.14)
0
a
Solution:
If a = 0 then (3.14) is trivial. Consider for simplicity a > 0, the case a < 0 is analogous. Recall
equation (2.26) from previous lecture
Z T
Z a+T
Z a
Z a+T
f (x) dx =
f (x) dx +
f (x) dx −
f (x) dx
0
a
0
T
Z a+T
Z a+T
Z a+T
=
f (x) dx +
f (x − T ) dx −
f (x) dx
a
T
T
Z a+T
Z a+T
Z a+T
=
f (x) dx +
f (x) dx −
f (x) dx
a
T
T
3-6
Exercise 3.8. Prove the following integration formulas, valid for β 6= 0:
Z x
1
cos(α + βt) dt = [sin(α + βx) − sin α]
β
Z0 x
1
sin(α + βt) dt = − [cos(α + βx) − cos α]
β
0
(3.15)
(3.16)
Solution:R
x
Consider 0 cos(α + βt) and use angle addition formula
Z x
Z x
(cos α cos βt − sin α sin βt)
cos(α + βt) =
0
0
Z βx
Z βx
1
1
= cos α
cos t − sin α
sin t
β
β
0
0
1
= [cos α sin βx − sin α(1 − cos βx)]
β
1
= [cos α sin βx + sin α cos βx − sin α]
β
1
= [sin(α + βx) − sin α]
β
The proof of (3.16) is analogous and left as an easy homework.
Exercise 3.9. Prove that for all integers n 6= 0
Z 2π
sin nx dx = 0
0
Z 2π
cos nx dx = 0
0
Solution:
Recall formula (2.26) from previous lecture.
Z
Z 2π
1 2nπ
sin x dx
sin nx dx =
n 0
0
1
= (1 − cos(2nπ))
n
=0
The proof of (3.18) is analogous and left as an easy homework.
(3.17)
(3.18)
3-7
Exercise 3.10. Using (3.17)-(3.18) and the addition formulas for sine and cosine establish the
following identities, valid for all integers m and n such that m2 6= n2 and n, m 6= 0.
Z 2π
sin nx cos mx dx = 0
(3.19)
0
Z 2π
sin nx sin mx dx = 0
(3.20)
0
Z 2π
cos nx cos mx dx = 0
(3.21)
0
Z 2π
sin2 nx dx = π
(3.22)
0
Z 2π
cos2 nx dx = π
(3.23)
0
Solution:
Recall the angle addition formulas for sine
sin(α + β) = cos α sin β + sin α cos β
sin(α − β) = − cos α sin β + sin α cos β
from which we obtain, subtracting the second equality from the first,
sin(α + β) − sin(α − β) = 2 sin β cos α
(3.24)
Therefore, by means of (3.24) and (3.17) we have
Z 2π
Z 2π
Z 2π
1
sin nx cos mx dx =
sin[x(n + m)] dx −
sin[x(n − m)] dx
2 0
0
0
=0
Recall now the angle addition formulas for cosine
cos(α + β) = cos α cos β − sin α sin β
cos(α − β) = cos α cos β + sin α sin β
from which we easily obtain
cos(α + β) − cos(α − β) = −2 sin α sin β
Therefore, by means of (3.25) and (3.18) we have
Z 2π
Z 2π
Z 2π
1
sin nx sin mx dx =
cos[(n − m)x] dx −
cos[(n + m)x] dx
2 0
0
0
=0
The proof of (3.21) is analogous and left as an easy homework.
(3.25)
3-8
Eventually, let us prove (3.22). From (3.25) we get
Z
2π
Z
2
sin nx dx =
0
2π
sin nx sin nx dx
0
Z 2π
Z 2π
1
cos[(n + n)x] dx
cos[(n − n)x] dx −
2 0
0
Z
1 2π
=
dx
2 0
=π
=
The proof of (3.23) is analogous and left as an easy homework.