Download Integration Involving Trigonometric Functions and Trigonometric

Integration Involving Trigonometric Functions
and Trigonometric Substitution
Dr. Philippe B. Laval
Kennesaw State University
September 7, 2005
Abstract
This handout describes techniques of integration involving various
combinations of trigonometric functions. It also describes a technique
known as trigonometric substitution. Students may want to review some
basic trigonometric identities before reading further. The following trigonometric identities will be used:
• sin2 x + cos2 x = 1
• 1 + tan2 x = sec2 x
1 − cos 2x
• sin2 x =
2
1
+
cos
2x
• cos2 x =
2
• sin 2x = 2 sin x cos x
In addition, students need to remember the following:
π
π
• sin x is invertible when − ≤ x ≤ . Its inverse is denoted sin−1 x.
2
2
π
π
• tan x is invertible when − < x < . Its inverse is denoted tan−1 x.
2
2
3π
π
or π ≤ x <
. Its inverse is
• sec x is invertible when 0 ≤ x <
2
2
−
1
denoted sec x.
1
Powers of Sine and Cosine
Before we explain
the technique,
let us recall that we can integrate integrals
of the form sinn x cos xdxor cosn x sin xdx where n is a positive integer, by
using substitution. For example, to integrate sinn x cos xdx, we let u = sin x
1
then du = cos xdx. Therefore
sinn x cos xdx = un du
un+1
n+1
sinn+1 x
=
n+1
=
The other integral is done similarly.
The technique used here depends on whether one of the powers is odd or
both are even. We summarize the techniques, then do some examples.
Proposition 1 Suppose we have an integral of the form sinm x cosn xdx.
1. If n is odd, that is n = 2k + 1, then save one cosine factor, and use the
identity sin2 x + cos2 x = 1 to express the remaining factors in terms of
sine. Then, use the substitution u = sin x. In other words
m
n
sin x cos xdx = sinm x cos2k+1 xdx
= sinm x cos2k x cos xdx
k
= sinm x 1 − sin2 x cos xdx
2. If m is odd, that is m = 2k + 1, then save one sine factor, and use the
identity sin2 x + cos2 x = 1 to express the remaining factors in terms of
cosine. Then, use the substitution u = cos x. In other words
sinm x cosn xdx = sin2k+1 x cosn xdx
= sin2k x cosn x sin xdx
k
=
1 − cos2 x cosn x sin xdx
3. If both m and n are even, we use the half-angle identities
1 − cos 2x
2
1 + cos 2x
2
cos x =
2
sin2 x =
as well as the identity
sin x cos x =
2
sin 2x
2
We illustrate the following techniques with some examples.
Example 2 Find cos3 xdx
This is the case where the power of cosine is odd. We save one cosine factor
and write
cos3 x = cos2 x cos x
= 1 − sin2 x cos x
Therefore,
1 − sin2 x cos xdx
= cos xdx − sin2 x cos xdx
cos3 xdx =
(1)
The first integral is known
cos xdx = sin x
(2)
The second integral can be evaluated using the substitution u = sin x =⇒ du =
cos xdx and therefore
(3)
sin2 x cos xdx = u2 du
u3
3
sin3 x
=
3
=
Using equation 2 and equation 3 in equation 1 gives us
sin3 x
+C
cos3 xdx = sin x −
3
Example 3 Find sin5 x cos2 xdx
This is the case where the power of sine is odd. We save one sine factor and
write
sin5 x cos2 x = sin4 x cos2 x sin x
2
= sin2 x cos2 x sin x
2
= 1 − cos2 x cos2 x sin x
= 1 − 2 cos2 x + cos4 x cos2 x sin x
= cos2 x sin x − 2 cos4 x sin x + cos6 x sin x
3
Therefore,
5
2
sin x cos xdx =
2
cos x − 2 cos4 x + cos6 x sin xdx
We then use the substitution u = cos x =⇒ du = − sin xdx to get
2
5
2
sin x cos xdx = −
u − 2u4 + u6 du
3
u
2u5 u7
=−
−
+
+C
3
5
7
cos3 x 2 cos5 x cos7 x
+
−
+C
=−
3
5
7
Example 4 Find sin2 xdx
This is the case when the powers of sine and cosine are even (the power of cosine
1 − cos 2x
being 0). We use the half angle identity sin2 x =
to obtain
2
1
2
(1 − cos 2x) dx
sin xdx =
2
We use the substitution u = 2x =⇒ du = 2dx to get
1
2
sin xdx =
(1 − cos u) du
4
1
= (u − sin u) + C
4
1
= (2x − sin 2x) + C
4
x sin 2x
+C
= −
2
4
Similar techniques can be applied to powers of tangent and secant. We will
not cover them here. They can be found in most Calculus books.
2
Trigonometric Substitution
The techniques we are about to describe apply to integrals containing expressions of the form
a2 − x2
a2 + x2
x2 − a2
for
which the other techniques have failed. For example, if we were given
√
x 1 − x2 dx, the substitution u = 1 − x2 would work. However, if we were
4
√
given
1 − x2 dx, it would be much more difficult to do. We will look at each
case separately. Before we do this, it is important to keep in mind an important
difference between the substitution technique learned before and the one we are
about to explain. In the traditional substitution, we define the new variable in
terms of the old. For example, u = 1 − x2 . In trigonometric substitution, we
redefine the given variable.
Remark 5 In order to be able to do this substitution successfully, you must be
able to find all the trigonometric functions, knowing one of them. This can be
done either by using trigonometric identities or a triangle. This technique can
be found in any book dealing with trigonometric functions. It can also be found
on the handout linked to on the web site for the class.
2.1
Integral Containing
√
a2 − x2
π
π
We use the substitution x = a sin θ, with − ≤ θ ≤
and a > 0. We impose
2
2
this restriction on θ so that sin θ will have an inverse. This substitution is based
on the identity 1 − sin2 θ = cos2 θ and works as follows:
x = a sin θ =⇒ x2 = a2 sin2 θ
=⇒ a2 − x2 = a2 − a2 sin2 θ
= a2 1 − sin2 θ
= a2 cos2 θ
Therefore
√
a2 − x2 = a2 cos2 θ
= |a cos θ|
= |a| |cos θ|
= a cos θ
We were able to remove the absolute value because a > 0 and cos θ ≥ 0 when
π
π
− ≤ θ ≤ . We illustrate this with examples.
2
2
√
9 − x2
dx
Example 6 Find
x2
π
π
We let x = 3 sin θ, with − ≤ θ ≤ . Then dx = 3 cos θ. Also, as noted above,
2
2
5
√
9 − x2 = 3 cos θ. Therefore,
√
9 − x2
dx
=
x2
=
=
=
3 cos θ
3 cos θdθ
9 sin2 θ
cos2 θ
dθ
sin2 θ
cot2 θdθ
2
csc θ − 1 dθ
= − cot θ − θ + C
We need to express our answer in terms of x. Since x = 3 sin θ, it follows that
x
θ = sin−1 . Also, either using trigonometric identities, or a triangle, we find
3 √
9 − x2
that cot θ =
. Therefore,
x
√
√
9 − x2
9 − x2
x
dx
=
−
− sin−1 + C
2
x
x
3
2√
Example 7 Find 0 4 − x2 dx
√
Method 1 We recognize that 4 − x2 is the upper half circle of radius 2 centered at the origin. The integral of it between 0 and 2 corresponds to the
area of the first quadrant of this circle. Therefore
2
1
4 − x2 dx = (area of a circle of radius 2)
4
0
1 2 =
2 π
4
=π
This method is very quick and easy. However, it would not work if the
problem had been to find an antiderivative. We show another technique,
using trigonometric substitution.
Method
√ 2 According to what was explained above, we let x = 2 sin θ. Then,
4 − x2 = 2 cos θ. Also, dx = 2 cos θdθ. To find the value of this integral,
we will first find an antiderivative, then use the given limits of integration.
Therefore,
2
4 − x dx = 4 cos θ cos θdθ
= 4 cos2 θdθ
6
Remembering the techniques of the previous section, we use cos2 θ =
Therefore,
1 + cos 2θ
4 − x2 dx = 4
dθ
2
= 2 (1 + cos 2θ) dθ
1 + cos 2θ
.
2
If we let u = 2θ, then du = 2dθ and we have
4 − x2 dx = (1 + cos u) du
= u + sin u
= 2θ + sin 2θ
= 2θ + 2 sin θ cos θ
We obtained the last equality using the identity sin 2θ = 2 sin θ cos θ. Now,
we write everything back in terms of x. First, since x = 2 sin θ, we see
that
x
sin θ =
2
and
x
θ = sin−1
2
To express cos θ in terms of x, we use cos2 θ = 1 − sin2 θ and since cos θ ≥
0, we have
cos θ = 1 − sin2 θ
x2
= 1−
4
4 x2
=
−
4
4
1
=
(4 − x2 )
4
√
4 − x2
=
2
Therefore
√
x 4 − x2
−1 x
2
4 − x dx = 2 sin
+2
2
2
2
√
x x 4 − x2
= 2 sin−1 +
2
2
7
We can now find the definite integral
2
√
2
2
x
4
−
x
x
−1
4 − x2 dx = 2 sin
+
2
2
0
0
√ √ 2
0 4
0
−1
−1
= 2 sin 1 +
− 2 sin 0 +
2
2
π
= 2 + 0 − (0 + 0)
2
=π
2.2
Integral Containing
√
a2 + x2
π
π
We use the substitution x = a tan θ, with a > 0 and − < θ < . We impose
2
2
this restriction on θ so that tan θ will have an inverse. The substitution is based
on the identity 1 + tan2 θ = sec2 θ and works as follows:
a2 + x2 = a2 + a2 tan2 θ
= a2 1 + tan2 θ
= a2 sec2 θ
Therefore
√
a2 + x2 = a2 sec2 θ
√
= a sec2 θ
= a |sec θ|
= a sec θ
π
π
Because a > 0 and sec θ ≥ 0 if − < θ < .
2
2
1
√
dx
Example 8 Find
x2 x2 + 4
√
We let x = 2 tan θ, dx = 2 sec2 θdθ. Also, x2 + 4 = 2 sec θ. Therefore:
2 sec2 θdθ
1
√
dx =
2
2
4 tan2 θ (2 sec θ)
x x +4
1
sec θ
=
dθ
4
tan2 θ
Now,
1
sec θ
cos
θ
=
tan2 θ
sin2 θ
cos2 θ
cos θ
=
sin2 θ
8
If we make the substitution u = sin θ, then du = cos θdθ and we get:
1
1
cos θ
√
dx =
dθ
2
2
4
sin2 θ
x x +4
1
du
=
4
u2
1
=
u−2 du
4
−1
=
+C
4u
−1
+C
=
4 sin θ
We express sin θ in terms of x and obtain
x
sin θ = √
4 + x2
Therefore
2.3
√
− 4 + x2
1
√
dx =
+C
4x
x2 x2 + 4
Integral Containing
√
x2 − a2
π
3π
We use the substitution x = a sec θ, with a > 0 and 0 ≤ θ < or π ≤ θ <
.
2
2
We impose this restriction on θ so that sec θ will be invertible. This substitution
is based on the identity sec2 θ − 1 = tan2 θ and works as follows:
x2 − a2 = a2 sec2 θ − a2
= a2 sec2 θ − 1
= a2 tan2 θ
Therefore
x2 − a2 = a2 tan2 θ
= |a| |tan θ|
= a tan θ
because a > 0 and tan θ ≥ o when 0 ≤ θ <
π
3π
or π ≤ θ <
.
2
2
dx
√
, where a > 0.
2
x − a2
According to the explanation above, we let x = a sec θ. Then, dx = a sec θ tan θdθ.
Example 9 Find
9
Also,
√
x2 − a2 = a tan θ. Therefore
dx
a sec θ tan θdθ
√
=
a tan θ
x2 − a2
= sec θdθ
= ln |sec θ + tan θ| + C
x
(see homework 1). Now, we need to write everything in terms of x. sec θ =
a
√
x2 − a2
and tan θ =
. Therefore,
a
x √x2 − a2 dx
√
= ln +
+C
2
2
a
a
x −a
= ln x + x2 − a2 − ln a + C
= ln x + x2 − a2 + C
3
Problems
1. Find sec θdθ. (hint: multiply both numerator and denominator by
sec θ + tan θ)
2. Find cos5 x sin5 xdx
3. Find cos4 xdx
√
a2 − x2 dx
4. Using the technique of example 7, find
x
dx
5. Find √
1 − x2
√
1 − 4x2 dx
6. Find
x
dx
7. Find √
2
x +3
√
8. Find et 9 − e2t dt
4
Answers
1.
sec θdθ = ln |sec θ + tan θ| + C
1 6
sin x −
6
3
1
3. cos4 xdx = x + sin 2x +
8
4
2.
cos5 x sin5 xdx =
1 8
1
sin x +
sin10 x + C
4
10
1
sin 4x
32
10
√
x√ 2
a2
x
a2 − x2 dx =
a − x2 +
sin−1 + C
2
2
a
√
x
dx = − 1 − x2 + C
5. √
1 − x2
4.
√
1
1 √
1 − 4x2 dx = sin−1 2x + x 1 − 4x2 + C
4
2
√
x
dx = x2 + 3 + C
7. √
x2 + 3
t
t√
1 t√
−1 e
2t
2t
8. e 9 − e dt =
e 9 − e + 9 sin
+C
2
3
6.
11