11.6 Hyperbolic Substitution
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Important note: Not all calculus classes teach hyperbolic
substitution. Each integrand addressed here can also be integrated
using variations of the trigonometric substitution method that were not
covered in the last section. We will be covering those in a later
section called More Substitutions, after we have covered
partial fractions. If hyperbolic substitution is not part of your
curriculum, then skip ahead (unless you're curious) to
Partial Fractions,
and don't worry, we will cover the methods to integrate the functions that are covered here
later on.
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In the last section we started with the problem of integrating
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dx
______
√1 - x2
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and discovered that in order to integrate this we needed to make a substitution,
f(u) = x, where
f(u) had
the very special property
_________
f'(u) = √1 - f2(u)
We found that
f(u) = sin(u) filled all the
requirements, and when we used it, it worked to find our indefinite integral in
this example. Indeed we later found that by
completing the square
we could use this same substitution to integrate a whole class of functions
that had a quadratic under the radical -- but not all such functions. It could
only do them if the
x2 term under the radical was negative
and the quadratic had real roots. So what happens if we try to integrate a
quadratic-under-the-radical that has neither of these properties?
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dx
______ eq. 11.6-1
√1 + x2
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In this case we would also like to make a substitution of
f(u) = x, but the property
of
f(u) that will make all the hard stuff cancel this
time is
_________
f'(u) = √1 + f2(u) eq. 11.6-2
If we could do that, then we could substitute
f(u) = x
and
f'(u) du = dx, to get
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_________
√1 + f2(u)
_________ du =
√1 + f2(u)
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du eq. 11.6-3
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But alas, no function that you have studied so far has the property described
in equation 11.6-2. And so we do what mathematicians do when they have the properties
of a function but no function in hand that meets those properties. They invent
one that does. Since the function we're after fills the same kind of need that the sine function did
in the last section, we'll name it after the sine function and call it
hyperbolic sine. The notation is
f(u) = sinh(u).
If you look carefully into what our hyperbolic sine function has to do, you
can infer a way of constructing it out of exponentials:
1
sinh(u) =
2
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(eu - e-u) eq. 11.6-4a
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How it is possible to discover this involves material to be covered in a later
section on differential equations,
so for now you can consider its derivation to be optional material. But if you're
curious about how to arrive at equation 11.6-4a based upon the property listed
in equation 11.6-2, click here.
By analogy we also define a hyperbolic cosine, which is the derivative
of hyperbolic sine.
1
cosh(u) =
2
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(eu + e-u) eq. 11.6-4b
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Notice that the way we have chosen these definitions, we have made
sinh(u) be an odd function, just as
sine is. We have made cosh(u) be an even
function, just as cosine is. And we have made
cosh(0) = 1 just like
cos(0). Although the derivative of the
hyperbolic sine is the hyperbolic cosine, the analogy
does not quite stretch to finding the derivative of the hyperbolic
cosine. Recall that the derivative of cos(x)
is -sin(x). But the derivative of the
hyperbolic cosine is once again the hyperbolic sine
with no minus sign. You can confirm this simply by taking
the derivative of the right-hand side of equation 11.6-4b and comparing
it to the right-hand side of equation 11.6-4a.
As you would expect, there is also a hyperbolic tangent function,
which we define analogously to the tangent function (which, like its,
trig analog is an odd function):
sinh(u)
tanh(u) = eq. 11.6-5
cosh(u)
whose derivative you can find using the
quotient rule:
dtanh(u)
=
du
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cosh2(u) - sinh2(u)
=
cosh2(u)
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1 - tanh2(u) eq. 11.6-6
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Notice the near analogy to what happens to the tangent function when you
take its derivative.
There are also hyperbolic cotangent (coth(u)),
hyperbolic secant (sech(u)), and
hyperbolic cosecant (csch(u)). These are
defined as you would expect based upon their trig analogs. We won't
discuss these three any further here because we don't need them in order to do
substitution on any of the integrals we will be studying.
Figure 11.6-1: Plots of Hyperbolic Functions
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You can see in these plots that the hyperbolic sine,
hyperbolic cosine,
and hyperbolic tangent functions look nothing like their trigonometric
analogs other than their behavior at x = 0.
Recall that sine and cosine were both bounded between -1 and 1.
Both hyperbolic sine and hyperbolic cosine are unbounded
(although hyperbolic cosine never goes less than 1). Nor
are hyperbolic sine and hyperbolic cosine periodic the way
sine and cosine are in the trig world. And to further contrast the hyperbolic
with the trig, note that the trig function, tangent, is unbounded and
has undefined points on its domain. But the hyperbolic tangent
is bounded to within a range between -1 and 1 and has a domain of all
the real numbers.
Despite all these differences, these functions provide the substitutions
we need to integrate a whole family of integrands that trig substitutions
are able to address only in a clumsy way. The integrands that the hyperbolic functions
address are all analogs of the ones that the trig functions were able
to address. And the substitution with hyperbolic functions is done analogously
as well.
As we had identities with trig functions, we have identities with
hyperbolic functions too. And you will see that all of their identities
are near analogies to corresponding trig identities.
BTW in conversation we have
traditional nicknames for the hyperbolic functions. Hyperbolic
sine is usually pronounced, "sinsh," like the word, "cinch,"
although I have heard it pronounced, "shin" also. Hyperbolic cosine
is pronounced, "cosh," as in the word, "kosher." And hyperbolic
tangent is pronounced, "tansh."
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Hyperbolic Identities
If you square each of the definitions in equations
11.6-4a and 11.6-4b and take
the second minus the first, you find:
1
4
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(e2u + 2 + e-2u) -
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1
4
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(e2u - 2 + e-2u) = 1 eq. 11.6-7a
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From that we can conclude
cosh2(u) - sinh2(u) = 1 eq. 11.6-7b
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for all values of u. Starting with that identity, you can
easily see that
____________
cosh(u) = √sinh2(u) + 1 eq. 11.6-8a
and
____________
sinh(u) = ±√cosh2(u) - 1 eq. 11.6-8b
where you take the plus of the ± when u is positive and
the minus when u is negative (Don't dwell too much on the ±
business in this identity. It turns out that for the purpose of integration, you
can almost always drop the ± when using this identity and still
end up getting the signs right in the end).
You can apply the sum-of-exponents rule and some factoring tricks to both
11.6-4a and 11.6-4b
to get sum rules for hyperbolic sine and hyperbolic cosine.
1
sinh(u+v) =
2
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(eu+v - e-u-v) =
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1
2
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(euev - e-ue-v) =
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1
4
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(euev - eue-v + e-uev - e-ue-v) +
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1
4
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(euev + eue-v - e-uev - e-ue-v) =
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1
4
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(ev - e-v)(eu + e-u) +
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1
4
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(eu - e-u)(ev + e-v)
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and out of that we find
sinh(u+v) = sinh(v)cosh(u) + sinh(u)cosh(v) eq. 11.6-9
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Likewise
1
cosh(u+v) =
2
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(eu+v + e-u-v) =
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1
2
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(euev + e-ue-v) =
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1
4
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(euev + eue-v + e-uev + e-ue-v) +
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1
4
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(euev - eue-v - e-uev + e-ue-v) =
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1
4
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(eu + e-u)(ev + e-v) +
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1
4
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(eu - e-u)(ev - e-v)
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From this we find
cosh(u+v) = cosh(u)cosh(v) + sinh(u)sinh(v) eq. 11.6-10
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From these you can get a sum formula for hyperbolic tangent
as well
sinh(u+v) sinh(u)cosh(v) + sinh(v)cosh(u)
tanh(u+v) = = =
cosh(u+v) cosh(u)cosh(v) + sinh(u)sinh(v)
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tanh(u) + tanh(v)
eq. 11.6-11
1 + tanh(u)tanh(v)
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Formulas for double values follow from the above:
sinh(2u) = 2sinh(u)cosh(u) eq. 11.6-12a
cosh(2u) = cosh2(u) + sinh2(u) eq. 11.6-12b
2tanh(u)
tanh(2u) = eq. 11.6-12c
1 + tanh2(u)
and from the double value formulas for hyperbolic sine and
hyperbolic cosine together with equation 11.6-7b we
get formulas for the squares of sinh(u) and
cosh(u):
sinh2(u) =
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1
2
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(cosh(2u) - 1) eq. 11.6-13a
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cosh2(u) =
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1
2
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(cosh(2u) + 1) eq. 11.6-13b
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Inverse Hyperbolic Functions
Since we were able to construct the hyperbolic functions from exponentials,
it shouldn't surprise you that we can construct their inverses from logs.
The easiest of them to analyze as a log is the hyperbolic arctangent.
If you have
eu - e-u
y = tanh(u) =
eu + e-u
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then you find
arctanh(y) by solving for
u
in terms of
y. I'll let you follow along with the algebra --
(eu + e-u)y = eu - e-u
e-u(e2u + 1)y = e-u(e2u - 1)
(e2u + 1)y = e2u - 1
y e2u + y = e2u - 1
1 + y = e2u - y e2u
1 + y = e2u (1 - y)
1 + y
= e2u
1 - y
1
ln
2
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1 + y
1 - y
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= u = arctanh(y) eq. 11.6-14
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Note that the domain of the hyperbolic arctangent is only
-1 < y < 1.
You can find the hyperbolic arcsine and hyperbolic arccosine in terms of
logs using the above together with
arcsinh(y) = arctanh
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y
______
√y2 + 1
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eq. 11.6-15a
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and
arccosh(y) = arctanh
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______
√y2 - 1
y
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eq. 11.6-15b
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See if you can use equations
11.6-8a and
11.6-8b
to verify for yourself why the above two equations work.
Hyperbolic arcsine has as its domain all the real numbers.
By constrast the
hyperbolic
arccosine has a domain only of real numbers whose absolute values are greater than or equal
to
1. Values returned by the above formula for
y < -1 are not strictly a result of its being
the inverse function of
hyperbolic cosine (see the
graph),
but that part of the formula's the domain is, nevertheless,
useful for evaluating certain integrals.
You can learn lots more about inverse hyperbolics by
clicking here.
Applying Hyperbolics to Integrals
When you have an integrand involving a quadratic under a radical, we have
seen that the first thing to do is complete the square.
The next thing is to factor the constant out of the radical.
On doing both of those, the radical will end up in one of the following
forms:
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a)
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__________
√1 - (x/a)2
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b)
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__________
√(x/a)2 + 1
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c)
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__________
√(x/a)2 - 1
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where
a is some positive constant.
At this point whatever part of the integrand cannot be subdued using
simple substitution you will attack by making
either a trig substitution or a hyperbolic substitution. Your choice of
substitution is based upon which of the
cases above the radical has arrived at. The substitutions for each of the cases are:
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a)
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x
sin(u) =
a
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and
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a cos(u) du = dx
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b)
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x
sinh(u) =
a
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and
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a cosh(u) du = dx
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c)
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x
cosh(u) =
a
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and
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a sinh(u) du = dx
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You have already seen in the last section how to employ trig identities in case
a).
In cases
b) and
c) you will be employing the hyperbolic identities we
have discussed in this section. These include equations
11.6-7b,
11.6-8a,
and
11.6-8b so that you can convert between
sinh(u)
and
cosh(u) (and vice versa). You will also make use of
the square formulas,
11.6-13a and
11.6-13b
as well as the double-value formulas,
11.6-12a and
11.6-12b. The use of all these formulas will be similar to the
way you used the analogous trig formulas when you did trig substitution.
Let's do an example.
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___________
√x2 - 6x + 5 dx eq. 11.6-16
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Step 1: Complete the square.
Using the method of adding half the middle coefficient to
x, we find
v = x - 3
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and
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v2 = x2 - 6x + 9
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To turn
v2 into the quadratic under the radical, you have
to subtract
4:
v2 - 4 = x2 - 6x + 5
Taking the derivative of
v = x - 3 yields that
dv = dx, and now we're ready to do the first
substitution:
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______
√v2 - 4 dv eq. 11.6-17a
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Step 2: Factor the constant from under the radical to get it into
one of the
standard forms. This results in
2
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__________
√(v/2)2 - 1 dv eq. 11.6-17b
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Step 3: Choose your next substitution.
Clearly the form you have at this point fits
standard form c).
The table tells you to substitute
v
cosh(u) =
2
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and
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2 sinh(u) du = dv
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That substitution results in
4
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____________
√cosh2(u) - 1 sinh(u) du eq. 11.6-17c
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According to the identity in equation
11.6-8b, this is
the same as
4
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(cosh2(u) - 1) du eq. 11.6-17d
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(I'm dropping the
± here, and we'll see that proper assignment
of signs works out by itself when we look at a graph of the result at the end of all
this). According
to equation
11.6-7b, we have
4
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sinh2(u) du eq. 11.6-17e
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One more identity to employ here to make this thing integrable. That is
equation
11.6-13a.
2
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(cosh(2u) - 1) du eq. 11.6-17f
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Step 4: Integrate. So what is the integral of the
hyperbolic cosine?
Since
hyperbolic sine and
hyperbolic cosine are each derivatives of
the other, it makes sense that they are also antiderivatives of each other. So
taking the integral above results in
2
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(cosh(2u) - 1) du =
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sinh(2u) - 2u + C eq. 11.6-18a
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Now use the
double value identity to convert that
to
sinh(2u) - 2u + C = 2sinh(u)cosh(u) - 2u + C eq. 11.6-18b
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Using equation
11.6-8b (again dropping the
±) we get:
____________
2sinh(u)cosh(u) - 2u + C = 2√cosh2(u) - 1 cosh(u) - 2u + C
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eq. 11.6-18c
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Step 5: Substitute back to v.
The
cosh(u) is replaced with
v/2.
The bare
u has to be replaced with
arccosh(v/2).
__________
v √(v/2)2 - 1 - 2 arccosh
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v
2
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+ C eq. 11.6-19a
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or
equivalently
1
2
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______
v √v2 - 4 - 2 arccosh
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v
2
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+ C eq. 11.6-19b
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Step 6: Substitute back to x.
Everywhere you see a
v, replace it with
x - 3.
1
2
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___________
(x - 3) √x2 - 6x + 5 - 2 arccosh
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x - 3
2
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+ C eq. 11.6-20
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Figure 11.6-2
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Here you can see plots of the function we just integrated along
with the integral we came up with, but with C = 0.
The integral is plotted exactly as the green expression shows it, but using
the extended domain for arccosh that we get from equation 11.6-15b.
There is no other manipulation of the signs.
Observe that both the original function and its integral are undefined
in the interval, 1 < x < 5. Observe
also that if you slid the whole graph 3 units left (which is the equivalent
of substituting v = x - 3), the blue function would
be even and the green function would be odd. That is how I know the signs
are right on the left-hand branch of the integral function. This is because the
integral of an even function is always a function that is odd to within the
addition of a constant. The symmetry of the graph tells me I got it right.
But how can you know that if you don't have a function plotter?
Observe that the integrand of equation 11.6-17a
is an even function. It represents the original integrand (the blue function)
slid 3 units to the left.
Equation 11.6-19b is the green function slid
3 units to the left. Note that its left term is an even times an odd function,
which is an odd function. The hyperbolic arccosine, as we defined it
in equation 11.6-15b is also odd. So the whole v
solution in equation 11.6-19b is an odd function. That
confirms that the signs, at that point, are consistent.
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Notice that the blue function in figure 11.6-2 is the conic section (or at least
part of it) that we call a hyperbola. That we use sinh and cosh
to integrate a hyperbola is why we call these functions hyperbolic.
Exercises
1) Find the indefinite integral of
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x dx
_____________
√x2 - 10x + 41
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Make the substitution that
completes the square.
Make sure you apply the corresponding substitution to the
x in the numerator.
Break the integral up into a sum. Decide which summand won't yield to
simple substitution.
On that summand, divide the factor out of the radical, then look up which form it is on the
table.
Make the substitution indicated by the
table to that summand.
Integrate and substitute everything back, and you'll have the solution. Then
rewrite it in logs if you like, using equation
11.6b-6, but that last step is not necessary.
See solution for exercise 1
2)
So far I've been picking coefficients for these problems carefully,
so that wherever you have to take the square root of some value, it turns
out that you're taking the square root of a perfect square. I can't guarantee
that all the problems on your exam will be so accommodating. To get you
ready for that eventuality, I'm going
to have you do one here where you don't have all perfect squares. Do keep
in mind that if one of these appears on an exam, you probably won't get
a warning like the one I'm giving you now. If that happens, check your
algebra (the most likely mistake at that point would be an error in completing the square).
It's up to you to determine whether the numbers not working out
nicely is due to your mistake or to your instructor's intentions. With
that in mind, find the indefinite integral of
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x2 dx
________
√7x2 - 21
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I figure you've completed enough squares by now that I don't have to drill
you on that one more time, so there is no need to complete the square in
this one. You will have to employ that last
trick
I showed you in the trig substitution section in order to prepare this one
for
polynomial long division.
You will have to divide out the factor under the radical, paying careful
attention to the algebraic rules for doing that. That will put the radical
into one of the standard forms for you to look up on the
table.
Make the substitution, integrate, and then substitute back.
See solution for exercise 2
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