Download MATH 118, LECTURE 8: Trigonometric Substitutions

MATH 118, LECTURE 8:
Trigonometric Substitutions
1
Trigonometric Substitution
In this lecture, we are going to use some of our trigonometric identities to
simplify integrals. The identities we use relate to the form of derivatives we
know: the derivatives of inverse trigonometric functions.
d
1
arcsin(x) = √
dx
1 − x2
d
1
arctan(x) =
dx
1 + x2
1
d
arcsec(x) = √
dx
x x2 − 1
⇐⇒ 1 − x2 ⇐⇒
1 − sin2 (x) = cos2 (x)
⇐⇒ 1 + x2 ⇐⇒
1 + tan2 (x) = sec2 (x)
⇐⇒ x2 − 1 ⇐⇒ sec2 (x) − 1 = tan2 (x).
Our approach will be a little different than in previous lectures. Rather than
simply integrating the above expressions, we will try a different intuition.
Consider the integral
Z
1
√
dx.
1 − x2
We recognize very quickly that we are looking at the integral of the derivative
of arcsin(x), so the fundamental theorem of calculus tells us that
Z
1
√
dx = arcsin(x) + C,
1 − x2
but let’s assume we are not quite so enlightened.
How could we solve the integral directly? We might as well start by
trying a substitution, but which substitution will work? The trouble point
is the square root, since it does not distribute over the terms contained
within it. We would much rather have a single term under the square root;
however, it can be easily checked that the substitution u = 1 − x2 will not
work. We appear to be stuck.
There is however, another substition which is capable of simplifying 1−x2
into a single term. The trick is to recall the trig identities
sin2 (x) + cos2 (x) = 1
1
(1)
1 + tan2 (x) = sec2 (x).
(2)
If we rearrange identity (1) we get cos2 (x) = 1 − sin2 (x), so if we choose
x = sin(u) (or u = arcsin(x)) then 1−x2 becomes 1−sin2 (u) which simplifies
to cos2 (u)!
Let’s try this:
Z
1
√
dx
1Z− x2
1
p
=
cos(u) du
1 − sin2 (u)
Z
cos(u)
du
=
| cos(u)|
Z
= 1 du = u + C = arcsin(x) + C.
This is exactly what we expected!
Note that I ignored the absolute value on the cos(x) in the denominator
when I cancelled the fraction. We are justified in doing this only because
the transformation x = sin(u) is defined over the range of its inverse, u =
arcsin(x), which is u ∈ [−π/2, π/2]. A fortunate consequence of this is that
cos(u) is never negative over u ∈ [−π/2, π/2], so that | cos(u)| = cos(u)
wherever the integral is defined.
So we can evaluate integrals containing troublesome terms of the form
1 − x2 by substituting x = sin(u). We can, in fact, handle more general
cases by adjusting by constants to get the following principle.
Proposition 1.1. To simplify integrals involving troublesome terms of the
form a2 − b2 x2 , try the substitution
a
b
x = sin(u), or u = arcsin
x
b
a
so that
a2 − b2 x2 = a2 cos2 (u).
Now consider
Z
1
dx.
1 + x2
We know already that the term to be integrated here is the derivative of
arctan(x), but let’s attempt to solve the problem from basic principles. The
problem here is the denominator 1 + x2 , but the substitution u = 1 + x2
2
does not work. We are left with trigonometric substitution as an option,
but which substitution should we use? The answer is readily apparent from
identity (2): if we choose x = tan(u) (or u = arctan(x)), we get 1+tan2 (u) =
sec2 (u). Our denominator has been simplified to a single term!
Combining everything, we have
Z
1
dx
1 + x2
Z
1
=
sec2 (u) du
1 + tan2 (x)
Z
sec2 (u)
=
du
sec2 (u)
Z
= 1 du = u + C = arctan(x) + C.
This is exactly what we expected! The substitution x = tan(u) is able to
simplify integrals containing troublesome 1 + x2 terms.
Again, we can adjust by constants so long as we can factor out a 1 + x2
term, so that we arrive at the following principle.
Proposition 1.2. To simplify integrals involving troublesome terms of the
form a2 + b2 x2 , try the substitution
a
b
x = tan(u), or u = arctan
x
b
a
so that
a2 + b2 x2 = a2 sec2 (u).
Lastly, consider directly solving
Z
1
√
dx.
x x2 − 1
Again, the troublesome term is the square root in the denominator.
The substitution u = x2 − 1 does not work, so we try a trigonometric
substitutions. We see that identity (2) can be rearranged into
tan2 (x) = sec2 (x) − 1
(3)
which suggests the transformation x = sec(u) (or u =arcsec(x)). This is
becoming a common theme - the substitution is the trigonometric function
whose derivative contains the troublesome term we are trying to simplify.
3
This leads to
Z
1
√
dx
x Zx2 − 1
1
p
=
sec(u) tan(u) du
sec(u) sec2 (u) − 1
Z
Z
sec(u) tan(u)
du = 1 du = u + C = arcsec(u).
=
sec(u) tan(u)
Again, we can adjust by constants to get the following principle.
Proposition 1.3. To simplify integrals involving troublesome terms of the
form b2 x2 − a2 , try the substitution
a
b
x = sec(u), or u = arcsec
x
b
a
so that
b2 x2 − a2 = a2 tan2 (u).
It should be clarified what I mean when I say troublesome terms. We
are only interested in trigonometric substitution (generally) when the terms
of interest (e.g. a2 + b2 x2 , etc.) appear in forms which cannot be directly
integrated (e.g. inside roots, denominators or fractions, etc.) and for which
direct substitution (e.g. u = a2 + b2 x2 , etc.) does not work. It is, in some
sense, our third line of defense. We try direct integration first. If that fails,
we try a direct substitution for the troublesome term. And if that does not
work, we should try a trigonometric substitution.
Example 1:
Evaluate
Z p
9 − 4x2 dx.
The square root prevents us from integrating directly and we are unable to
simplify the integral using u = 9 − 4x2 , so we are left with trigonometric
substitution. We pick x = (3/2) sin(u) (or u = arcsin((2/3)x)). Intuitively,
we have picked these constants so that the 2 will cancel the 4 and the 3 will
factor with the 9 (after squaring).
4
We have
Z p
9 − 4x2 dx
s
2
Z
3
3
9−4
=
sin(u) cos(u) du
2
2
Z q
3
9 − 9 sin2 (u) cos(u) du
=
2
Z q
9
=
1 − sin2 (u) cos(u) du
2
Z
9
cos2 (u) du
=
2
Z
9
1 + cos(2u)
=
du
2
2
9
sin(2u)
=
u+
+C
4
2
9
= (u + sin(u) cos(u)) + C
4
9
2
9
2
2
= arcsin
x + sin arcsin
x
cos arcsin
x
+C
4
3
4
3
3
!
√
9
2
9 2
9 − 4x2
= arcsin
x +
x
+C
4
3
4 3
3
2
1 p
9
x + x 9 − 4x2 + C.
= arcsin
4
3
2
This example illustrates how involved integrating by trigonometric substitution can be. We have had to make use of several identities (cos2 (x) =
(1/2)(1 + cos(2x)) and sin(2x) = 2 sin(x) cos(x)). If it were not for these
identities, we would have not been able to solve the problem so it is important to know these identities, and how to apply then, well.
We also note that, after substituting back to the original variable (in
this case x), we will be expected to construct triangles to solve for terms
like cos(arcsin(x)), sin(arctan(x)), etc.
Example 2:
Evaluate
Z
x2
1
dx.
−1
5
Again, we cannot integrate directly, and the substitution u = x2 −1 does
not work, so we try a trigonometric substitution. The denominator dictates
we use the substitution x = sec(u) (or u =arcsec(x)). We have
Z
1
dx
2
x −1
1
=
sec(u) tan(u) du
2
sec (u) − 1
Z
sec(u)
du
=
tan(u)
Z
= csc(u) du
= − ln | csc(arcsec(u)) + cot(arcsec(u))| + C
x
1
+C
= − ln √
+√
x2 − 1
x2 − 1 x+1 +C
= − ln √
x2 − 1 √
x2 − 1 = ln + C.
x+1 The solution suggests another substitution which would have worked,
namely,
√
x2 − 1
u=
.
x+1
For completeness, let’s see if we can solve the integral using this substitution
instead.
(x2 − 1)−1/2 x(x + 1) − (x2 − 1)1/2
dx
du =
(x + 1)2
=⇒ dx =
(x2 − 1)1/2 (x + 1)2
du
x(x + 1) − (x2 − 1)
(x2 − 1)1/2 (x + 1)2
du
x+1
= (x2 − 1)1/2 (x + 1) du
=
6
We have
Z
1
dx
x2 − 1
Z
(x2 − 1)1/2 (x + 1)
du
=
x2 − 1
Z
x+1
=
du
2
(x − 1)1/2
√
Z
x2 − 1 1
=
du = ln |u| + C = ln + C.
x+1 u
So this substitution would also have worked... if we had been able to
find it! It is, however, well beyond our intuition to guess such a substitution prior to first solving the integral using trigonometric substitution. The
second substitution also did not lead to significantly less work. Regardless,
this shows there is often more than one way to approach these problems.
Example 3:
Evaluate
Z
x(x2
1
dx
+ 3)1/2
We check our identities and see √
very quickly that this fits the
√ form of a
tan(u) substitution. We have x = 3 tan(u) (or u = arctan(x/ 3)). This
7
gives us
Z
1
dx
+ 3)1/2
√
Z
3 sec2 (u)
√
=
du
3 tan(u)(3 tan2 (u) + 3)1/2
Z
1
sec(u)
=√
du
tan(u)
3
1
= √ csc(u) du
3
1
= √ ln |csc(u) − cot(u)| + C
3 1
x
x
+C
= √ ln csc arctan √
− cot arctan √
3
3
3
√
x2 + 3 √3 1
= √ ln −
+C
x
x 3 √
x2 + 3 − √3 1
= √ ln + C.
x
3 x(x2
Example 4:
Evaluate
Z
1−x
dx.
(3 − x2 )1/2
At first glance, this may look like our previous examples; however, the x
term in the numerator makes it subtle difference. Remember that trigonometric substitution is our third line of defense. We cannot integrate any
portion of this expression directly, but the substitution u = 3 − 4x2 will
work on the second term if we expand the integral as
Z
Z
Z
1−x
1
x
dx =
dx −
dx.
1/2
1/2
2
2
(3 − x )
(3 − x )
(3 − x2 )1/2
√
√
We will use the substitution x = 3 sin(u1 ) (or u1 = arcsin((1/ 3)x)) for
8
the first integral and u2 = 3 − x2 in the second to get
Z
1−x
dx
(3 − x2 )1/2
Z
Z
1
x
=
dx −
dx
1/2
2
(3 − x )
(3 − x2 )1/2
Z √
Z
x
3 cos(u1 )
1
√
=
du1 −
−
du2
1/2
2x
3 cos(u1 )
u2
Z
Z
1
−1/2
= 1 du1 +
u2
du2
2
1/2
= u1 + u2 + C
p
1
√
= arcsin
x + 3 − x2 + C.
3
Application 1:
Of course, we would not be learning about trigonometric substitution if it
did not allow us to solve problems of broader interest. For a simple example,
let’s consider calculating the area of a circle of radius r. We have been told
since elementary school that the Area=πr2 , but can we demonstrate this
using calculus? Let’s check.
We recall that the definite integral gives us the area under the curve
bound by two points, so consider a circle centred at (0, 0). It will extend
along the x-axis a magnitude of r in both directions, so our bounds of
integration are x from −r to r. Also, since the circle is symmetric about the
x-axis, we can just integrate the top half and then double the result!
The equation of a circle of radius r is
x2 + y 2 = r 2
which we can solve for in terms of y to get
p
y = r 2 − x2 .
This gives the upper half of the circle, which is all we need. Putting everything together, we have
Z rp
Area = 2
r2 − x2 dx.
−r
9
We recognize this immediately as the form for a sin(u) substitution. We
use the substitution x = r sin(u), which gives dx = r cos(u) du. We have
Z rp
2
r2 − x2 dx
−r
Z x=r q
= 2r
r2 − r2 sin2 (u) cos(u) du
Zx=−r
x=r
cos2 (u) du
= 2r2
x=−r
Z x=r
[1 + cos(2u)] du
= r2
x=−r
sin(2u) x=r
2
=r u+
2
x=−r
= r2 [u + sin(u) cos(u)]x=r
x=−r
"
#
x x √r 2 − x2 r
2
= r arcsin
+
r
r
r
−r
= r2 [arcsin(1) − arcsin(−1)]
h π π i
− −
= πr2 .
= r2
2
2
We should feel relieved that our elementary school teachers were not
lying to us. Notice that I could have substituted the bounds in terms of u
as well, and solved directly. We notice that at x = −r, x = r sin(u) implies
u = arcsin(−1) = −π/2, and at x = r we have u = arcsin(1) = π/2, so that
(skipping intermediate steps)
Z rp
2
r2 − x2 dx
−r
= ···
π/2
= r2 [u + sin(u) cos(u)]−π/2
π π π −π
−π
2 π
=r
+ sin
cos
− −
− sin
cos
2
2
2
2
2
2
π i
h
2 π
2
=r
− −
= πr
2
2
2
Trigonometric Integrals (Revisisted!)
There is one more class of simple trigonometric integrals which we can handle
by making use of trigonometric identities. The identities we are interested
10
in are:
sin((n + m)x) + sin((n − m)x)
2
cos((n − m)x) − cos((n + m)x)
sin(nx) sin(mx) =
2
cos((n + m)x) + cos((n − m)x)
cos(nx) cos(mx) =
2
Notice that we cannot integrate the terms on the left, but the terms on
the right are easily integrated.
sin(nx) cos(mx) =
Example 1:
Evaluate
Z
sin(5x) cos(3x) dx.
We cannot integrate directly, but we can use our identities to get
Z
sin(5x) cos(3x) dx
Z
1
[sin(8x) + sin(2x)] dx
=
2
cos(8x) cos(2x)
1
−
−
+C
=
2
8
2
cos(8x) cos(2x)
=−
−
+ C.
16
4
Example 2:
Evaluate
Z
√
sin(3x) sin( 7x) dx.
Again, we use our identities to get
Z
√
sin(3x) sin( 7x) dx
Z
i
√
√
1 h
=
cos((3 − 7)x) − cos((3 + 7)x) dx
2
"
#
√
√
1 sin((3 − 7)x) sin((3 + 7)x)
√
√
=
−
+ C.
2
3− 7
3+ 7
11