Download Problem Set 4 Partial Solutions

Math 142.009, Problem Set 4 Solutions (Partial)
Due Wednesday, Feb. 9, 2011
From Text Calculus: Concepts and Contexts, 3rd edition, by James Stewart
Section 5.6, pp. 398-399, # 1, 4, 17, 18, 21, 25, 28, 37, 39
Section 5.7, pp. 404-405, # 1, 2, 4, 5, 9, 10
Additional Problems: Evaluate the following indefinite integrals:
Z
1.)
x3 ln x dx
We integrate by parts, which means that we apply the integration by parts formula
Z
Z
(1)
u dv = uv − v du.
Because it is easier to differentiate ln x that to integrate it, we let u = ln x. Then since the integral we are
R
1
1
doing must be of the form u dv, we let dv = x3 dx. Then du = dx and v = x4 . Using equation (1),
x
4
we get
Z
Z
Z
1 1 4
1
1
1
· x dx = x4 ln x −
x3 dx
x3 ln x dx = x4 ln x −
4
x 4
4
4
1
1 1
1
1
= x4 ln x − · x4 + C = x4 ln x − x4 + C.
4
4 4
4
16
2.)
Z
x sec2 x dx
d
tan x = sec2 x. Hence it seems reasonable to integrate by parts, letting u = x and
dx
dv = sec2 dx. Then du = dx and v = tan x. Hence by equation (1), we get
Z
Z
Z
sin x
2
x sec x dx = x tan x − tan x dx = x tan x −
dx.
cos x
Z 0
d
f
Since
cos x = − sin x, the last integral is of the form
dx, so
dx
f
Z
sin x
dx = ln | cos x| + C,
−
cos x
Recall that
(formally, you can do this integral by making the substitution y = cos x). Hence
Z
x sec2 x dx = x tan x + ln | cos x| + C.
3.)
Z
y 2 e−y dy
We integrate by parts, letting u = y 2 and dv = e−y dy. Then du = 2y dy and v = −e−y . Hence using
equation (1),
Z
Z
Z
2 −y
2 −y
−y
2 −y
y e dy = −y e − −e 2y dy = −y e + 2ye−y dy.
We do the last integral by integrating by parts again. Let u = 2y and dv = e−y dy. Then du = 2 dy and
v = −e−y . Hence
Z
Z
Z
−y
−y
−y
−y
2ye dy = −2ye − −2e dy = −2ye + 2 e−y dy = −2ye−y − 2e−y + C.
Putting this together, we get
Z
4.)
Z
y 2 e−y dy = −y 2 e−y − 2ye−y − 2e−y + C.
sin3 x cos4 x dx
Since one of the trig functions is raised to an odd power, we can reduce it to a single power by using
the identity sin2 x + cos2 x = 1. Then sin2 x = 1 − cos2 x, so
Z
Z
Z
3
4
2
4
sin x cos x dx = sin x sin x cos x dx = sin x(1 − cos2 x) cos4 x dx.
Now we make the substitution u = cos x. Then du = − sin x dx. Hence
Z
Z
Z
Z
2
4
2 4
4
6
sin x(1 − cos x) cos x dx = − (1 − u )u du = − u − u du = u6 − u4 du
1
1
1
1
= u7 − u5 + C = cos7 x − cos5 x + C.
7
5
7
5
5.)
Z
cos4 x dx
Since the trigonometric function in this integral is raised to an even power, we must use the trigonometric identity
1 1
(2)
cos2 θ = + cos 2θ.
2 2
Using (2) with θ = x gives
2
Z
Z
Z 2
1 1
4
2
cos x dx =
cos x dx =
+ cos 2x dx
2 2
Z
1
1
1
+ 2 · cos 2x + cos2 x dx
=
4
4
4
1
=
4
Z
1 + 2 cos 2x + cos2 2x dx.
1
1
The last integral includes the term cos2 2x, so we use (2) with θ = 2x to write cos2 2x = + cos 4x.
2
2
Hence
Z
Z
1
1
1 1
1 + 2 cos 2x + cos2 2x dx =
1 + 2 cos 2x + + cos 4x dx
4
4
2 2
Z
3
1
1
+ 2 cos 2x + cos 4x dx
=
4
2
2
1
1 3
x + sin 2x + sin 4x + C
=
4 2
8
3
1
1
= x + sin 2x +
sin 4x + C.
8
4
32
6.)
Z
x2
√
1
dx
4 − x2
√
We make the trigonometric substitution x = 2 sin θ (because of the term 4 − x2 in the integral). Then
dx = 2 cos θ dθ and
4 − x2 = 4 − 4 sin2 θ = 4(1 − sin2 θ) = 4 cos2 θ.
√
Hence 4 cos2 θ = 2 cos θ, so
Z
Z
1
1
√
dx =
2 cos θ dθ
2
4 sin θ · 2 cos θ
x2 4 − x2
Z
Z
1
1
1
csc2 θ dθ
=
2 dθ =
4
4
sin θ
1
1 cos θ
= − cot θ + C == −
+ C.
4
4 sin θ
To put the answer back in terms of x, we draw a right triangle which represents our substitution x = 2 sin θ,
or sin θ = x/2, with side opposite the angle θ labelled √x and the hypoteneuse labelled 2. Then by the
Pythagorean theorem the side adjacent to θ has length 4 − x2 :
2
√
θ
x
4 − x2
Hence
cos θ =
Therefore
Z
√
4 − x2
2
and
sin θ =
1
1 cos θ
1
√
dx = −
+C =− ·
2
2
4 sin θ
4
x 4−x
√
4 − x2
=−
+ C.
4x
√
x
.
2
4 − x2 2
· +C
2
x