Download Homework 2 Solutions

Math 2315 - Calculus II
Homework #2 - 2007.08.31
Due Date - 2007.09.07
Solutions
Part 1: Problems from sections 7.3 and 7.4.
Section 7.3:
19.
Z √
1 + x2
dx
x
We will use the substitution x = cot(θ), dx = − csc2 (θ). This gives
Z √
Z p
1 + x2
1 + cot2 (θ)
dx = −
csc2 (θ)dθ.
x
cot(θ)
Next, some algebraic manipulations:
Z p
Z
1 + cot2 (θ)
2
−
csc (θ)dθ = −
cot(θ)
Z
=−
Z
=−
Z
=−
Z
=−
tan(θ) csc3 (θ)dθ
csc2 (θ) csc(θ) tan(θ)dθ
1 + cot2 (θ) csc(θ) tan(θ)dθ
csc(θ) tan(θ) + csc(θ) cot(θ)dθ
Z
sec(θ)dθ + − csc(θ) cot(θ)dθ.
Both of the final integrals above we can do. So we get
Z
Z
− sec(θ)dθ + − csc(θ) cot(θ)dθ = − ln (sec(θ) + tan(θ)) + csc(θ) + D.
We now have to draw a triangle. Since x = cot(θ),
we get that tan(θ) = x1 (we
√
√
2
could do that without the triangle), and sec(θ) = xx+1 , csc(θ) = x2 + 1. So we
now have
!
√
Z √
p
2
2
1+x
x +1 1
dx = − ln
+
+ x2 + 1 + D.
x
x
x
1
27.
Z
dx
(x2 + 2x + 2)2
First, we complete the square: gives
Z
Z
dx
dx
=
2
2
2
(x2 + 2x + 2)
(x + 1) + 1
and then use the substitution u = x + 1, which gives
Z
Z
du
dx
.
2 =
2 + 1)2
2
(u
(x + 1) + 1
Now, we use the trigonometric substitution u = tan(θ):
Z
Z
du
sec2 (θ)
=
2 dθ
(u2 + 1)2
(tan2 (θ) + 1)
Z
sec2 (θ)
dθ
=
sec4 (θ)
Z
dθ
=
sec2 (θ)
Z
= cos2 (θ)dθ
Z
1 1
=
+ cos(2θ)dθ
2 2
1
1
= θ + sin(2θ) + D.
2
4
Next, we need to get back to the variable x, so first, we use the identity: sin(2θ) =
2 sin(θ) cos(θ), this gives
1
1
1
1
θ + sin(2θ) + D = θ + sin(θ) cos(θ) + D.
2
4
2
2
Now, since u = tan(θ), we can draw a triangle to find both sin(θ) and cos(θ), and
we find that sin(θ) = √uu2 +1 and cos(θ) = √u12 +1 . So
1
1
1
1 u
θ + sin(θ) cos(θ) + D = tan−1 (u) +
+ D.
2
2
2
2 u2 + 1
Finally, substituting u = x + 1 gives
Z
du
1
1
x+1
= tan−1 (x + 1) +
+ D.
2
2
2 (x + 1)2 + 1
(u2 + 1)
2
Section 7.4:
25.
Z
10
dx
(x − 1)(x2 + 9)
So partial fractions it is. We guess
A
Bx + C
10
=
+
,
(x − 1)(x2 + 9) x − 1
x2 + 9
and can solve for A by setting x = 1. This gives A = 1. So now the expression is
10 = x2 + 9 + Bx2 − Bx + Cx − C.
Since this has to hold for all x, this forces B = −1 (why?) and hence C = −1 as
well (again why?). So
Z
Z
10
1
x+1
dx
=
−
dx.
(x − 1)(x2 + 9)
x − 1 x2 + 9
We now expand:
Z
Z
Z
Z
1
x+1
1
x
1
− 2
dx =
dx −
dx
−
dx.
x−1 x +9
x−1
x2 + 9
x2 + 9
Each one of the above integrals can be done without too much trouble. So we have
Z
1
1
10
2
−1 x
dx = ln(x − 1) − ln(x + 9) − tan
+ R.
(x − 1)(x2 + 9)
2
3
3
47.
e2x
dx
e2x + 3ex + 2
Here we set u = ex , and du = ex dx, so we have
Z
Z
Z
e2x
ex
u
x
dx
=
e
dx
=
du.
e2x + 3ex + 2
e2x + 3ex + 2
u2 + 3u + 2
The right-most integral above is in the form for partial fractions. So we set:
Z
u
A
B
=
+
,
u2 + 3u + 2 u + 2 u + 1
and find that B = −1 and A = 2 by setting u = −1 and u = −2 respectively. So
now
Z
Z
u
2
1
du
=
−
du = 2 ln(u + 2) − ln(u + 1) + D.
u2 + 3u + 2
u+2 u+1
3
After resubstitution, we get
Z
e2x
dx = 2 ln(ex + 2) − ln(ex + 1) + D.
2x
x
e + 3e + 2
4
Part 2: The fun problems.
1. Consider the following integral:
Z
t5
√
dt.
t2 + 2
a) Solve the integral in two ways, first by using the substitution u =
and then solving the resulting trigonometric integral.
√
2 tan(u)
√
Using the substitution suggested, we have dt = 2 sec2 (u)du and
√
Z
Z
t5
( 2)5 tan5 (u) √
√
2 sec2 (u)du
dt =
√ p 2
2
t +2
2 tan (u) + 1
√ 5 Z tan5 (u) sec2 (u)
= ( 2)
du
sec(u)
√ 5Z
tan5 (u) sec(u)du
= ( 2)
√ 5Z
= ( 2)
tan4 (u) sec(u) tan(u)du
√ 5Z
2
= ( 2)
sec2 (u) − 1 sec(u) tan(u)du
Now we are ready to set v = sec(u) and dv = sec(u) tan(u)du. So we have
√ 5Z
√ 5Z
2
2
2
( 2)
sec (u) − 1 sec(u) tan(u)du = ( 2)
v 2 − 1 dv
√ 5Z 4
v − 2v 2 + 1dv
= ( 2)
√ 5 1 5 2 3
= ( 2)
v − v + v + D.
5
3
Now we start back substituting. First up was that v = sec(u). So
√ 5 1 5 2 3
√ 5 1
2
( 2)
v − v + v + D = ( 2)
sec5 (u) − sec3 (u) + sec(u) + D.
5
3
5
3
√
And finally, we have to use the fact that t = 2 tan(u). Or more precisely, √t2 =
5
√
2
t +2
tan(u). Drawing a triangle gives that sec(u) = √
. So our final answer is
2
"
#
√
Z
5
3
2
2
5
2
√
2
2
1 (t + 2)
2 (t + 2)
t
t +2
√
√
√
dt = ( 2)5
−
+ √
+D
5 ( 2)5
3 ( 2)3
2
t2 + 2
p
5
3
1
4
= (t2 + 2) 2 − (t2 + 2) 2 + 4 t2 + 2 + D.
5
3
b) Solve the integral by the method of integration by parts. It might be helpful
to rewrite the integral as follows:
Z
t
dt,
t4 √
t2 + 2
and let u = t4 and v 0 =
√ t .
t2 +2
So we follow the suggested first step. If u = t4 and v 0 = √t2t+2 , then u0 = 4t3 and
√
v = t2 + 2. So we have
Z
Z
p
p
t
t4 √
dt = t4 t2 + 2 − 4t3 t2 + 2dt
t2 + 2
Z
p
p
4
2
2
= t t + 2 − 4 t t t2 + 2dt.
Notice that the integral on the second√line has been put into the form for another
3
substitution. Here, u = t2 and v 0 = t t2 + 2, making u0 = 2t and v = 13 (t2 + 2) 2 .
This gives
Z p
Z
p
p
3
3
1
2
4
3
4
2
2
2
t t2 + 2 − 4 t t2 + 2dt = t t2 + 2 − 4 t (t + 2) 2 −
t(t + 2) 2 dt
3
3
Z
p
3
3
8
4
= t4 t2 + 2 − t2 (t2 + 2) 2 +
t(t2 + 2) 2 dt.
3
3
Now the last integral is simple to do, so
Z
p
t
t4 √
dt = t4 t2 + 2 −
t2 + 2
we now have
3
5
4 2 2
8
t (t + 2) 2 + (t2 + 2) 2 + D.
3
15
c) Your two answers, which should both be in terms of the original variable t.
If you have done everything correctly, your answer to part a) should look different
than your answer to part b). Show that you answer to part a) is indeed equal to
your answer to part b).
6
We now have to compare the following two functions:
p
3
5
4
8
f1 = t4 t2 + 2 − t2 (t2 + 2) 2 + (t2 + 2) 2 + D
3
15
and
p
5
3
1
4
f2 = (t2 + 2) 2 − (t2 + 2) 2 + 4 t2 + 2 + D.
5
3
√
The simplest way to do this is by direct calculation. I.e. we will factor a t2 + 2
out of each function. After doing this, the first becomes
√
2 4
2 2
2
2
t + 2 15 t − 20 t t + 2 + 8 t + 2
f1 =
15
and the second
√
2
2
2
2
t + 2 3 t + 2 − 20 t + 20
f2 =
.
15
Next, we notice that
2
2
15 t4 − 20 t2 t2 + 2 + 8 t2 + 2 = 3 t2 + 2 − 20 t2 + 20 = 3t4 − 8t2 + 32.
So finally, we have that
√
t2 + 2
f1 = f2 =
3t4 − 8t2 + 32 .
15
2. Consider the following relationship between x and z:
x
z = tan
2
Show that under the above relation, one has
(
2
cos(x) = 1−z
1+z 2
2z
sin(x) = 1+z
2.
Hint: Using double angle identities will help!
So here, we use the hint:
(
cos(x) = 2 cos2 x2 − 1
sin(x) = 2 sin x2 cos x2 .
7
Consider the first identity:
cos(x) = 2 cos
2
x
2
2
1 + tan2
1 − z2
=
.
1 + z2
=
−1=
x
2
2
sec2
−1=
x
2
−1
2
−1
1 + z2
Now the second:
sin x2
2 x
sin(x) = 2 sin
cos
=1
· cos
2
2
2
cos x2
x
2 tan x2
1
=
= 2 tan
2 sec2 x2
1 + tan2 x2
2z
.
=
1 + z2
x
x
3. Using problem 2, show the following:
Z
x
1
dx = tan
+C
1 + cos(x)
2
Using problem 2, let z = tan x2 . Then cos(x) is given as above. However, we
must compute dx.
2dz
.
x = 2 tan−1 (z) =⇒ dx =
1 + z2
Thus, one has
Z
Z
Z
x
1
1
2dz
dx =
= dz = z + c = tan
+ C.
1−z 2 1 + z 2
1 + cos(x)
2
1 + 1+z
2
8