Math 1172 (Buenger)
1.
Z
Section 7.4
We use the Pythagorean
√ theorem to show
2
that the hypotenuse
√ is 4 + x . With this
in mind sec θ = 4 + x2 /2 and tan θ = x/2.
Thus
Z
dx
1 √
√
= ln 4 + x2 /2 + x/2 +C.
2
2
16 + 4x
dx
√
16 + 4x2
Solution:
Z
Z
dx
dx
√
√
=
.
2
16 + 4x
2 4 + x2
After the above simplification we observe
that the term under the radical is of the
form a2 + x2 , with a=2. Thus the substitution x = 2 tan θ, with dx = 2 sec2 θ dθ,
will simplify the integral.
Z
2.
Z
14
12
x3
√
dx
x2 − 100
Solution: The term under the radical is
of the form variable squared less constant
squared. Thus, based on the identity sec2 θ−
1 = tan2 θ, we should make the substitution
x = 10 sec θ, with dx = 10 sec θ tan θ dθ.a
Z
dx
√
dx
3 x2 − 100
x
Z
10 sec θ tan θ dθ
√
=
3 θ 100 sec2 θ − 100
1000
sec
Z
tan θ dθ
=
100 sec2 θ10 tan θ
Z
1
cos2 θ dθ
=
1000
Z
1
1 + cos(2θ)
=
·
dθ
1000
2
1
sin(2θ)
θ+
+ C.
=
2000
2
2 sec2 θ
p
dθ
2 4 + (2 tan θ)2
Z
2 sec2 θ
√
dθ
=
2θ
4
sec
Z
1
sec θ dθ
=
2
1
=
ln |sec θ + tan θ| + C.
2
dx
√
=
2 4 + x2
September 23, 2015
Z
Now we just need to find sec θ and tan θ in
terms of x. Consider the reference trianglea
formed by theta.
To finish we need to know what sin(2θ) is in
terms of x.
a
Recall, that one must be extra careful when
performing trig substitution √
with x = a sec θ, as
it is not always true that sec2 θ − 1 = tan θ.
When θ ∈ [0, π/2] or x ∈ [1, ∞), we have that
√
sec2 θ − 1 = tan θ. When
√ θ ∈ [π/2, π] or x ∈
(−∞, −1], we have that sec2 θ − 1 = − tan θ. In
this case we consider x values in [12, 14]. Thus we
will want to take the positive square root.
In our case tan θ = x2 , so we will let the adjacent leg of our triangle be 2 and the opposite
leg of our triangle be x.
a
By reference triangle we mean a right triangle
with theta as one of the smaller angles.
1
By the identity
Thus
Z √
sin(2θ) = 2 sin θ cos θ,
this amounts to finding sin θ and cos θ. By
definition, we know that cos θ = 10
and by
x
the following comparison triangle,
Z
=
Z
=
Z
=
Z
=
1 − x2
x
p
1 − sin2 θ
cos θ dθ
sin θ
√
cos2 θ
cos θ dθ
sin θ
cos2 θ
dθ
sin θ
1 − sin2 θ
dθ
sin θ
Z
√
=
2
it follows that sin θ = x x−100 . Thus
Z
dx
√
dx
3
x x2 − 100
10√x2 −100 x
sec−1 10
+
x2
=
+ C.
2000
csc(θ) − sin θ dθ
= − ln | csc θ + cot θ| + cos θ + C.
To finish we need to find what csc θ and cot θ
are in terms of x. To do this, we construct
the reference triangle below.
So
Z
14
x3
12
sec−1
=
sec
dx
dx
x2 − 100
10√142 −100 14
+
10
142
√
−1
−
3.
2000
+
12
10
√
10 122 −100
122
2000
Z √
.
Since x = sin θ we will let the hypotenuse
of our right triangle be 1 and the opposite side be x. By the Pythagorean
theo√
rem, the √remaining leg is 1 − x2 . Thus
cos
θ = 1 − x2 , csc θ = x1 , and cot θ =
√
1−x2
. Thus
x
1 − x2
dx
x
Z √
Solution: The term under the radical is of
the form a2 − x2 with a = 1. Thus we need
to make the substitution x = sin θ, with
dx = cos θ dθ.
2
1 − x2
x
√
dx = ln x
1 + 1 − x2 √
+ 1 − x2 + C.
4.
Z
1
4
5.
(Test question) Find an appropriate
trigonometric substitution of one of the forms
x = C sin(θ), x = C sec(θ), or x =
R C tan(θ)
4
to simplify the following integral: √xx2 −5 dx.
Leave your answer as an INDEFINITE integral of trig. functions of θ. (Simplify
this integral so as to eliminate the square
root, but do NOT evaluate the integral.)
Solution: In this case, the term under the
radical is x2 −5 indicating that our substitution should be of the form x = C sec θ. We
must√choose C to balance out the 5. Here
C = 5. Let
√
x = 5 sec θ.
1
dx
2
x − 2x + 10
Solution: Let us complete the square in the
denominator:
x2
1
1
=
.
− 2x + 10
(x − 1)2 + 9
Let x = 1 + 3 tan u and dx = 3 sec2 udu.
Then
Z
1
dx
2
x − 2x + 10
Z
1
dx
=
(x − 1)2 + 9
Z
3 sec2 u
=
du
(1 + 3 tan u − 1)2 + 9
Z
3 sec2 u
=
du
9 tan2 u + 9
Z
3 sec2 u
=
du
9 sec2 u
Z
1
du
=
3
1
=
u+C
3
1
x−1
−1
=
tan
+ C.
3
3
Then
dx =
√
5 sec θ tan θ dθ.
After substituting our integral becomes:
Z
=
√
4
5 sec θ
q √
2
5 sec θ − 5
25 sec4 θ
p
5 (sec2 θ − 1)
Z
25 sec4 θ
=
√ √ 2
5 tan θ
Z
25 sec4 θ
√
=
.
5 tan θ
Z
=
3
6. (Test Question) Find an appropriate 7. (Test Question) True or False:
trigonometric substitution of one of the forms
Z
Z
x4
tan4 θ
x = C sin θ, x = C sec θ, or x = C tan θ to
dθ
dx
=
(x2 + 3)5/2
sec3 θ
simplify the integral given below. Simplify
the integrand as much as possible but DO
√
where
x
=
3 tan θ.
NOT EVALUATE THE INTEGRAL.
√
True:
If
x
=
3 tan θ, then dx =
√
3 sec2 θ dθ, and
Z
Z
x4
x4
dx
2
5/2
dx
(2 + x )
(x2 + 3)5/2
√
Z
Solution: In this case, the term under the
√
( 3 tan θ)4
2
2
radical is 2 + x indicating that our substi=
√
5/2 3 sec θ dθ
2
( 3 tan θ) + 3
tution should be of the form x = C tan θ.
√
5
We must choose√C to balance√out the 2, and
3 tan4 θ sec2 θ
dθ
=
so we let
√ 5
√ C = 2. Let x = 2 tan θ. Then
3 sec5 θ
dx = 2 sec2 θ dθ, and after substituting
Z
tan4 θ
our integral becomes:
dθ.
=
sec3 θ
Z
x4
dx
(2 + x2 )5/2
√
4
Z
√
2 tan θ
=
2 sec2 θ dθ
5
√
2 2
2+
2 tan θ
Z
5
2 2 tan4 θ sec2 θ
=
5 dθ
(2 (1 + tan2 θ)) 2
Z 5
2 2 tan4 θ sec2 θ
dθ
=
5
(2 (sec2 θ)) 2
Z
tan4 θ
=
dθ.
sec3 θ
4