Math 222: Homework 1
Solutions1
§1.2: 2,3,4
§1.4: 1,3,4,7,8,9,11,12
§1.2
2. In both integrals A and B, the variable of integration is the same as the
independent variable of the function being integrated. Thus,
Z 4
A=B=
x−2 dx
1
4
1
= (− x−3 )
3
1
1
1
=− +
64 3
61
=
192
In integral C, on the other hand, the variable of integration is not the same as the
independent variable of the integrand. We therefore treat the independent
variable of the function as a constant and obtain
Z 4
4
C=
x−2 dx = (tx−2 ) = 4x−2 − x−2 = 3x−2 .
1
1
3a.
4
Z
(1 − x2 ) dx = (x −
2
=−
x3 4
)
3 2
50
3
<0
Thus the given inequality is false.
3b. Since the integral is with respect to t, we must treat the variable x in the
integrand as a constant. We therefore obtain
Z 4
4
(1 − x2 ) dt = (t(1 − x2 )) = 2(1 − x2 ).
2
2
Since there is no additional information about the variable x, we cannot determine
if this is greater than zero or not. The given inequality is therefore false.
3c. Observe that
Z
(1 + x2 ) dx = x +
x3
+ C.
3
For no value of the constant C is the function x +
conclude that the given inequality is false.
1
x3
3
Prepared by Kostas Beros, email: kberos@math.wisc.edu
1
+ C strictly positive. We
§1.4
1. Note that the second line below uses the formula sin2 α = 21 (1 − cos 2α).
Z
Z
(1 + sin 2θ)2 dθ = (1 + 2 sin 2θ + sin2 2θ) dθ
Z
1
= (1 + 2 sin 2θ + (1 − cos 4θ)) dθ
2
1
1
= θ − cos 2θ + θ − sin 4θ + C
2
8
1
3
= θ − cos 2θ − sin 4θ + C
2
8
3. We will use the following two identities:
1
sin 2α
2
1
sin2 α = (1 − cos 2α).
2
These justify the second and third lines below, respectively.
Z
Z
sin2 x cos2 x dx = (sin x cos x)2 dx
Z
1
= ( sin 2x)2 dx
2
Z
1
(1 − cos 4x) dx
=
8
1
1
= x−
sin 4x
8
32
sin α cos α =
4. In the second line below, we make the substitution u = sin θ (hence
du = cos θ dθ).
Z
Z
cos5 θ dθ = cos θ sin4 θ dθ
Z
= u4 du
1 5
u +C
5
1
= sin5 θ + C
5
=
7. Recall the formula
sin α cos β =
1
(sin(α + β) + sin(α − β)).
2
It follows that
Z
π
1
2
Z
π
(sin 5x + sin x) dx
1
1 π
= (− cos 5x − )
10
2 0
6
=
5
sin 3x cos 2x dx =
0
0
8. Recall the following three identities:
1
(1 + cos 2α)
2
1
sin2 α = (1 − cos 2α)
2
1
sin α cos β = (sin(α + β) + sin(α − β)).
2
First expand the integrand and then use these formulas to re-write the result:
Z
Z
(sin 2θ − cos 3θ)2 dθ = (sin2 2θ − 2 sin 2θ cos θ + cos2 3θ) dθ
Z
1
1
1
= ( (1 − cos 4θ) − (sin 3θ + sin θ) + (1 + cos 6θ)) dθ
2
2
2
1
1
1
1
= θ − sin 4θ + cos 3θ + cos θ +
sin 6θ
8
6
2
12
cos2 α =
9. Recall the following identity:
1
(cos(α − β) − cos(α + β)).
2
Using this as well as the half-angle formulas from the last problem, we re-write
the given integral as follows:
Z π2
Z π2
2
(sin 2θ + sin 4θ) dθ =
(sin2 2θ + sin 2θ sin 4θ + sin2 4θ) dθ
sin α sin β =
0
0
Z π
1 2
=
(1 − cos 4θ + cos 2θ − cos 6θ + 1 + cos 8θ) dθ
2 0
π2
1
1
1
1
1
= (2θ − sin 4θ + sin 2θ − sin 6θ − sin 8θ)
2
4
2
6
8
0
π
=
2
11a. Using the formula
sin α cos β =
1
(sin(α + β) + sin(α − β)),
2
we obtain
Z π
Z π2
1 2
sin(aθ) cos θ dθ =
(sin((a + 1)θ) + sin((a − 1)θ) dθ
2 0
0
π2
1
1
1
= (−
cos((a + 1)θ) −
cos((a − 1)θ))
2 a+1
a−1
0
1
1
π
1
π
= (−
cos((a + 1) ) −
cos((a − 1) )
2 a+1
2
a−1
2
1
1
+
+
)
a+1 a−1
1
1
1
1
π
1
π
= (
+
+
cos(a ) −
cos(a ))
2 a+1 a−1 a+1
2
a−1
2
a
1
π
= 2
+
sin(a )
a − 1 a2 − 1
2
The second to last line follows from the angle addition and subtraction formulas
for cosine applied to the terms
π
cos((a + 1) )
2
and
π
cos((a − 1) ).
2
11b. Use the double angle formula
sin θ cos θ =
1
sin 2θ
2
to obtain
Z
0
π
2
1
sin θ cos θ dθ =
2
Z
π
2
sin 2θ dθ
0
π2
1
= − cos 2θ
4
0
1
=
2
12. Again, we use the identity
sin α sin β =
1
(cos(α − β) − cos(α + β)).
2
to obtain
Z
Z t
1 t
sin(t − s) sin(as) ds =
(cos(t − (1 + a)s) + cos(t − (1 − a)s)) ds
2 0
0
t
1
1
1
sin(t − (1 + a)s) −
sin(t − (1 − a)s))
= (−
2 1+a
1−a
0
1
1
1
= (−
(sin(−as) − sin t) −
(sin(as) − sin t))
2 1+a
1−a
1
1
1
1
1
= ((
−
) sin(at) + (
+
) sin t)
2 1+a 1−a
1+a 1−a
a
1
=(
sin(at) +
sin t)
2
1−a
1 − a2
Note that the second to last line uses the fact that sin(−α) = − sin α and the last
line uses the the following two algebraic manipulations
1
1
2a
−
=
1+a 1−a
1 − a2
1
1
2
+
=
1+a 1−a
1 − a2