Math 2414
Homework Set 3 – Solutions
10 Points
#1. (2 pts) We first need to get the arguments on the trig functions the same. Changing the sine is best.
⌠ sin ( 4t )
⌠ 2sin ( 2t ) cos ( 2t )
⌠ 2sin ( 2t )
dt 
dt  =
dt
=
=
2
2
 cos
cos ( 2t )
( 2t )
⌡
⌡
⌡ cos ( 2t )
tan ( 2t ) dt
∫ 2=
ln sec ( 2t ) + c
#3. (2 pts) To start this one off we’ll use the substitution u = ln x to write the integral as,
⌠
4 − ln x
4−u
4−u
⌠
⌠=
=
dx
du
du



2
 x ln x 2 − 7 ln x − 30
⌡
−
−
−
+
u
u
u
u
7
30
10
3
(
)(
)
⌡
]
⌡ [
(
)
Now we need to partial fraction this.
4−u
A
B
=
+
( u − 10 )( u + 3) u − 10 u + 3
⇒
4 − u= A ( u + 3) + B ( u − 10 )
Plugging u = −3 into the numerators gives B = − 137 and plugging in u = 10 gives A = − 136 . So,
⌠
4 − ln x
1
6
7
1
6 ln ln ( x ) + 7 + ln ln ( x ) + 3 + c
dx =
− ⌠
+
du =−


 x ln x 2 − 7 ln x − 30
⌡
13
10
3
13
u
u
−
+
]
⌡ [
(
(
)
)
#7. (2 pts) First set the integral up as a limit and then note that the problem point is at x = 2 .
4
t
4
4
3
3
3
3
⌠
⌠
⌠
⌠
dx =
dx lim− 
dx + lim+ 
dx
=
 2
t →2
t →2
⌡1 x + x − 6
⌡ 1 ( x + 3)( x − 2 )
⌡ 1 ( x + 3)( x − 2 )
⌡ t ( x + 3)( x − 2 )
Now, we’ll need to partial fraction the integral and that’s a fairly simple partial fraction so I’ll leave it to
you to verify that,
3
3⌠ 1
1
3
⌠
=
dx
−
=
dx

( ln x − 2 − ln x + 3 )

5⌡ x−2 x+3
5
⌡ ( x + 3)( x − 2 )
Now, let’s just look at the each limit/integral above.
t
t
3
⌠
lim− 
dx =lim− 53 ( ln x − 2 − ln x + 3 ) =lim− 53 ( ln t − 2 − ln t + 3 − ln 4 ) =
−∞
1
t →2
t →2
t →2
⌡ 1 ( x + 3)( x − 2 )
4
4
3
⌠
∞
lim+ 
dx =lim+ 53 ( ln x − 2 − ln x + 3 ) =lim+ 53 ( ln t + 3 − ln t − 2 + ln 2 − ln 7 ) =
t
t →2
t →2
t →2
⌡ t ( x + 3)( x − 2 )
So, both integrals diverge and so the original integral also diverges (recall it only requires one to get
divergent for the original integral). Also, do NOT cancel the infinities and there is no reason to think that
they cancel here and the integral still diverges…..
Math 2414
Homework Set 3 – Solutions
10 Points
#8. (2 pts) In this case we’ve got an integral that is different from any others. We have an infinite limit
as well as division by zero at z = 2 so this is a mix of the two types. We’ll need to split up the integral
into one with a problem in each. It doesn’t matter where we split the integral so we’ll use z = 0 .
−4
−5
−4
−5
z +4
e z +4
e z +4
e z +4
e z +4
⌠
⌠
⌠
⌠
⌠ e
dz
dz
dz
dz
dz
=
+
=
lim
+
lim





z +4
z +4
t →−∞ ⌡
t →−4 − ⌡ 1 − e
⌡ −∞ 1 − e z + 4
⌡ −∞ 1 − e z + 4
⌡ −5 1 − e z + 4
t 1− e
−5
t
Now, let’s do the integral.
z +4
1
⌠ e
dz =
−⌠
− ln 1 − e z + 4
 du =

z +4
⌡u
⌡ 1− e
u=
1 − e z +4
Now let’s do each of the limits/integrals above.
−5
z +4
(
e
lim ⌠
dz = lim − ln 1 − e z + 4

z+4
t →−∞ ⌡
t →−∞
t 1− e
)
−5
t
(
= lim ln 1 − et + 4 − ln 1 − e −1
t →−∞
)
ln 1 − ln 1 − e −1 =
=
− ln 1 − e −1
e z +4
lim− ⌠
dz = lim− − ln 1 − e z + 4

z +4
t →−4 ⌡ 1 − e
t →2
−5
t
(
)
−5
t
(
= lim− ln 1 − e −1 − ln 1 − et + 4
t →2
)
=ln 1 − e −1 − ln 1 − 1 =∞
So, the original integral is divergent because the second integral is divergent (the fact that the first is
convergent is meaningless here).
#10. (2 pts) In this case the trig functions don’t add much to the function so looking only at the powers
of x in the integrand we can guess this will probably diverge and so we’ll need a smaller function we
know or can prove diverges.
9 x 2 + sin 2 ( x )
9x2 + 0
≥
x 3 − 2 cos 4 ( x + 4 ) x 3 − 2 cos 4 ( x + 4 )
9x2
9
=
≥ 3
x − 2 ( 0) x
Now we know that
∫
∞
1
9 x
b/c
0 ≤ sin 2 ( x ) ≤ 1
b/c
0 ≤ cos 4 ( x + 4 ) ≤ 1
dx diverges because p = 1 ≤ 1 and so by the Comparison Test we know that
9 x 2 + sin 2 ( x )
⌠
 x 3 − 2 cos 4 x + 4 dx
(
)
⌡9
∞
must also diverge.
Math 2414
Homework Set 3 – Solutions
10 Points
Not Graded
#2. First expand out the integrand.
∫ ( 2sin ( t ) − 6t )
2
dt=
∫ 4sin ( t ) − 24t sin ( t ) + 36t
2
2
∫ 36t
dt=
2
+ 2 (1 − cos ( 2t ) ) − 24t sin ( t ) dt
With a rewrite the first two terms can be integrated directly. We’ll need to integrate by parts on the
dv = sin ( t ) dt
third using : u = t
∫ ( 2sin ( t ) − 6t )
du = dt v = − cos ( t )
(
dt= 12t 3 + 2t − sin ( 2t ) − 24 −t cos ( t ) + ∫ cos ( t ) dt
2
)
= 12t 3 + 2t − sin ( 2t ) + 24t cos ( t ) − 24sin ( t ) + c
( )
#4. In this case we’ll use the trig substitution and note that e 2 z = e z
=
ez
1
4
tan θ
e z=
dz
1
4
sec 2 θ dθ
∫e
4z
16e 2 z =
+ 1 dz
∫e
=
1
256
3z
16e 2 z + 1=
e z dz
∫ ( sec
sec θ
Now, a right triangle give us=
2
∫(
1
4
( )
and e 4 z = e z
4
+ 1 sec=
θ sec θ
tan 2 θ =
+1
16e 2 z =
The integral is then,
2
tan θ ) ( sec θ ) ( 14 sec 2 θ ) dθ
3
θ − 1) sec 2 θ tan θ sec θ dθ=
1
256
(
1
5
sec5 θ − 13 sec3 θ ) + c
16e 2 z + 1 and so the integral is,
4z
2z
+ 1 dz
∫ e 16e =
1
256
5
3
1

2z
2 − 1 16e 2 z + 1 2 + c
16
+
1
e
(
)
(
)
5

3


#5. First rewrite the integral and then it’s a really simply substitution.
∫e
4 y + e4 y
u
4y e
1
1 e
+c
dy =
4 ∫ e du =
4e
∫ e e dy =
4y
4y
e4 y
u=
#6. First get it set up as a limit and then we’ll need to do integration by parts on the integral.
∞
t
3
3
⌠ ln ( 2 x )
⌠ ln ( 2 x )
dx = lim 
dx

4
t →∞
x4
⌡3
⌡3 x
u = ln ( 2 x3 )
du =
3
x
dv = x −4 dx
v = − 13 x −3
Math 2414
Homework Set 3 – Solutions
10 Points
3
 ln ( 2 x3 )

 ln ( 2 x3 ) 1 
⌠ ln ( 2 x )
−4


−
dx
lim
x
dx
lim
=
−
+
=
− 3

∫
t →∞ 
t →∞ 

x4
3x3
3x3
3x 
⌡3

3

3
t
∞
t
 ln ( 54 ) 1 ln ( 2t 3 ) 1 
= lim 
+ −
− 3
t →∞ 
81
81
3t 3
3t 


=
ln ( 54 ) 1
 3  ln ( 54 ) 1
 1 
+ − lim  t =
+ − lim  =


2
81
81 t →∞  9t 
81
81 t →∞  3t 3 
ln ( 54 ) 1
+
81
81
L’Hospital’s Rule shows up pretty regularly with these problems so don’t forget to use it where needed!
So, the integral is convergent and its’ value is,
∞
3
⌠ ln ( 2 x )
=
dx

x4
⌡3
ln ( 54 ) 1
+
81
81
#9. The numerator will never be that large and the exponential in the denominator will quickly become
very large and so it looks like this should converge so we’ll need a new, larger, function that we know or
can prove converges.
8 + 3sin 2 ( 9 x ) 8 + 3 (1)
=
≤
11e −2 x
2x
2x
e
e
I’ll leave it to you to verify that
∫
∞
2
b/c
0 ≤ sin 2 ( 9 x ) ≤ 1
11e −2 x dx converges and so by the Comparison Test we also know
that
∞
2
⌠ 8 + 3sin ( 9 x )
dx

e2 x
⌡2
must also converge.
#11. For this problem we have f ( x ) = e
1−sin ( x )
1
and ∆ x =
.
2
MidPoint Rule
∫
1
−1
1−sin ( x )
e
6.17651274
dx ≈ 12  f ( −0.75 ) + f ( 0.25 ) + f ( 0.25 ) + f ( 0.75 )  =
Trapezoid Rule
∫
1
−1
e1−sin ( x ) dx ≈ 14  f ( −1) + 2 f ( −0.5 ) + 2 f ( 0 ) + 2 f ( 0.5 ) + f (1)  =
6.2652477
Simpson’s Rule
∫
1
−1
e1−sin ( x ) dx ≈ 16  f ( −1) + 4 f ( −0.5 ) + 2 f ( 0 ) + 4 f ( 0.5 ) + f (1)  =
6.201304138
For comparison’s sake,
Math 2414
Homework Set 3 – Solutions
∫
2
0
cos (1 + e x ) dx =
6.206366177
10 Points