MATH 104 HW 8
CLAY SHONKWILER
§7.1
6. Evaluate the integral
Z
π/3
π/4
sec2 z
dz.
tan z
Answer: Let u = tan z. Then du = sec2 zdz, so the above integral
becomes
Z √3
√
√
1
du
= ln |u|]1 3 = ln 3 − ln 1 = ln 3.
u
2
1
35. Evaluate
Z eπ/3
dx
.
x cos(ln x)
1
Answer: Let u = ln x. Then du = x1 dx, so the above integral becomes
Z π/3
Z π/3
du
1 √
1 √
π/3
=
sec udu = ln | sec u + tan u|]0 = ln
+ 3 −ln(1) = ln
+ 3 .
cos u
2
2
0
0
38. Evaluate
4
Z
2
2dx
.
x2 − 6x + 10
Answer: Note that
(x2 − 6x) + 10 = (x2 − 6x + 9) + 10 − 9 = (x − 3)2 + 1,
so we can re-write the integral as
Z 4
2
2dx
.
(x − 3)2 + 1
Now, letting u = x − 3, this in turn is equal to
Z 1
−1 1
π −π
du
=
2
2
=
2
tan
u
−
= π.
−1
2
2
4
4
−1 u + 1
44. Evaluate
Z
(csc x − tan x)2 dx.
Answer: First, note that
Z
Z
2
(csc x − tan x) dx = (csc2 x − 2 csc x tan x + tan2 x)dx.
1
2
CLAY SHONKWILER
Now,
R
csc xdx = − cot x + C1 . Further,
1
sin x
1
·
=
= sec x
sin x cos x
cos x
R
R
and sec x = ln | sec x + tan x| + C2 , so 2 csc x tan x = 2 ln | sec x + tan x| +
C2 . Finally,
csc x tan x =
Z
Z
2
tan xdx =
(sec2 x − 1)dx = tan x − x + C3 .
Thus, if C = C1 + C2 + C3 ,
Z
(csc x − tan x)2 dx = cot x + ln | sec x + tan x| + tan x − x + C.
48. Evaluate
x2
dx.
x2 + 1
Z
Answer: Using long division,
x2
dx =
x2 + 1
Z
Z 1−
1
dx = x − tan−1 x + C.
x2 + 1
55. Evaluate
π/4
Z
0
1 + sin x
dx.
cos2 x
Answer: Splitting the numerator, we see that
Z
π/4
0
Z π/4
1
sin x
dx +
dx
2
cos x
cos2 x
0
0
Z π/4
Z
2
=
sec xdx + sec x tan xdx
1 + sin x
dx =
cos2 x
Z
π/4
0
π/4
= tan x + sec x]0
√ = 1 + 2 − (0 + 1)
√
= 2
58. Evaluate
Z
1
dx.
1 + cos x
MATH 104 HW 8
3
Answer: Note that we can multiply by a form of 1 in the following way:
Z
Z
1
1
1 − cos x
dx =
·
dx
1 + cos x
1 + cos x 1 − cos x
Z
1 − cos x
=
dx
1 − cos2 x
Z
1 − cos x
dx
=
sin2 x
Z
Z
1
cos x
=
dx
2 dx −
sin x
sin2 x
Z
Z
= csc2 xdx − csc x cot xdx
= − cot x + csc x + C.
83.
R
(a): Evaluate cos3 θdθ.
Answer: Note that
Z
Z
Z
Z
3
2
cos θdθ = (1 − sin θ) cos θdθ = cos θdθ − sin2 θ cos θdθ.
R
Now, cos θdθ = sin θ + C1 . Further, if u = sin θ, then du = cos θdθ,
so we can re-write the second integral as
Z
u3
sin3 θ
u2 du =
+ C2 =
+ C2 .
3
3
Hence, if C = C1 + C2 ,
Z
sin3 θ
cos3 θdθ = sin θ +
+ C.
3
R
(b): Evaluate cos5 θdθ.
Answer: Note that cos5 θ = cos2 θ cos3 θ = (1 − sin2 θ) cos3 θ, so
Z
Z
Z
cos5 θdθ = cos3 θdθ − sin2 θ cos3 θdθ.
R
Now, we already computed cos3 θdθ in (a). On the other hand,
Z
Z
Z
Z
sin2 θ cos3 θdθ = sin2 θ(1−sin2 θ) cos θdθ = sin2 θ cos θdθ− sin4 θ cos θ.
3
5
In each case, we let u = sin θ, so this integral becomes sin3 θ − sin5 θ +
C1 . Hence, combining the above, we see that
3
Z
sin3 θ
sin θ sin5 θ
sin5 θ
5
cos θdθ = sin θ +
−
−
+ C = sin θ +
+ C.
3
3
5
5
(c): Without
R actually evaluating the integral, explain how you would
evaluate cos9 θdθ.
4
CLAY SHONKWILER
Answer: We would proceed as in the above cases, making substitutions for cos2 θ and using what we know about the integrals of
cos to lower powers.
§7.2
7. Evaluate
Z
tan−1 ydy
1
Answer: Let u = tan−1 y and dv = dy. Then du = 1+y
2 dy and v = y, so
Z
Z
ydy
tan−1 ydy = y tan−1 y −
.
1 + y2
Now letting u = 1 + y 2 , du = 2ydy, so
Z
Z
ydy
1
du
1
1
=
= ln |u| + C1 = ln 1 + y 2 + C1 .
2
1+y
2
u
2
2
Therefore,
Z
tan−1 ydy = y tan−1 y −
1
ln 1 + y 2 + C.
2
10. Evaluate
Z
x3 ex dx.
Answer: Let u = x3 and dv = ex dx. Then du = 3x2 dx and v = ex , so
Z
Z
3 x
3 x
x e dx = x e − 3 x2 ex dx.
Now, letting u = x2 and dv = ex dx, du = 2xdx and v = ex , so
Z
Z
2 x
2 x
x e dx = x e − 2 xex dx.
Next, letting u = x and dv = ex , we see that du = dx and v = ex , so
Z
Z
x
x
xe dx = xe − ex dx = xex − ex .
Therefore, putting it all together,
Z
x3 ex dx = x3 ex − 3 x2 ex − 2 (xex − ex ) = x3 ex − 3x2 ex + 6xex − 6ex + C.
16. Evaluate
Z
t2 e4t dt.
Answer: Let u = t2 and dv = e4t dt. Then du = 2tdt and v =
Z
Z
4t
1
2 4t
2e
t e dt = t
−
te4t dt.
4
2
e4t
4 .
Hence,
MATH 104 HW 8
5
4t
Letting u = t and dv = e4t dt, du = dt and v = e4 . Thus,
Z
Z
e4t 1
e4t e4t
4t
te dt = t
−
e4t dt = t
−
.
4
4
4
16
Thus,
Z
4t
t2 e4t te4t e4t
1 e4t e4t
2 4t
2e
t e dt = t
t
=
−
−
−
+
+ C.
4
2
4
16
4
8
32
22. Evaluate
Z
e−y cos ydy
Answer: Let u = cos y and dv = e−y dy. Then du = − sin ydy and
v = −e−y , so
Z
Z
e−y cos ydy = −e−y cos y − e−y sin ydy.
Now, let u = sin y and dv = e−y dy. Then du = cos y and v = −e−y dy, so
Z
Z
e−y sin ydy = −e−y sin y + e−y cos ydy.
Therefore,
Z
Z
Z
−y
−y
−y
−y
−y
−y
e cos ydy = −e cos y− −e sin y + e cos ydy = −e cos y+e sin y− e−y cos ydy.
Adding
R
e−y cos ydy to both sides, we see that
Z
2 e−y cos ydy = e−y sin y − e−y cos y + C1 ,
so, if C = C1 /2,
Z
sin y − cos y
e−y sin y − e−y cos y
+C =
+ C.
e−y cos ydy =
2
2ey
26. Evaluate
Z 1
√
x 1 − xdx.
0
√
Answer: Let u = x and dv = 1 − xdx. Then du = dx and v =
− 32 (1 − x)3/2 . Hence,
1
Z 1
Z
√
2 1
2
3/2
x 1 − xdx = − x(1 − x)
+
(1 − x)3/2 dx
3
3
0
0
0
1
2
2 2
3/2
5/2
= − x(1 − x) − · (1 − x)
3
3 5
0
4
= (0 − 0) − 0 −
15
4
= .
15
6
CLAY SHONKWILER
27. Evaluate
Z
π/3
x tan2 xdx.
0
Answer: Note that
Z π/3
Z π/3
Z
2
2
x tan xdx =
x(sec x − 1)dx =
0
Now,
R π/3
0
xdx =
0
i
2 π/3
x
2
0
2
=
π2
18 .
xdx.
On the other hand, if u = x and dv =
π/3
tan xdx
0
π√
π/3
3 − [ln | sec x|]0
3
π√
3 − (ln 2 − ln 1)
=
3
√
π
=
3 − ln 2.
3
Therefore,
0
π/3
0
=
Z
Z
x sec xdx −
0
sec2 xdx, then du = dx and v = tan x, so
Z π/3
Z
π/3
= x tan x]0 −
0
π/3
π/3
x tan2 xdx =
π√
π2
3 − ln 2 − .
3
18
DRL 3E3A, University of Pennsylvania
E-mail address: [email protected]