3)
The example is to integrate
|
dx
_______
x2 √x2 - 16
|
First we have to divide out the constant. But with the
x2
sitting outside the radical, we have to divide that constant out twice
so that we end up with both
x2's in the same form
when we're
done.
|
dx
_______ =
x2 √x2 - 16
|
1
64
|
|
dx
__________ eq. 11.8-38
(x/4)2 √(x/4)2 - 1
|
We learned
earlier in this section
that the trig substitution to make here is
x
sec(u) =
4
|
and
|
___________
4 sec(u) √sec2(u) - 1 du = dx eq. 11.8-39a
|
which yields
1
64
|
|
dx
__________ =
(x/4)2 √(x/4)2 - 1
|
1
16
|
|
___________
sec(u) √sec2(u) - 1
___________ du = eq. 11.8-39b
sec2(u) √sec2(u) - 1
|
1
16
|
|
1
cos(u) du = sin(u) + C
16
|
The actual integration was easy enough to include in the above equation.
Now we have to substitute back. For that we take
the last entry in our recent
table
of trig identities
___________
√sec2(u) - 1
sin(u) =
sec(u)
|
When we back-substitute to
x,
we get
|
dx
_______ =
x2 √x2 - 16
|
1
16
|
__________
√(x/4)2 - 1
+ C =
x/4
|
1
16
|
_______
√x2 - 16
+ C eq. 11.8-40
x
|
Now we'll start again at equation 11.8-38 and
use hyperbolic substitution to arrive at the same result. The
hyperbolic substitution table tells
us to substitute
x
cosh(u) =
4
|
and
|
____________
4 sinh(u) du = 4 √cosh2(u) - 1 du = dx eq. 11.8-41a
|
which yields
1
64
|
|
dx
__________ =
(x/4)2 √(x/4)2 - 1
|
1
16
|
|
____________
√cosh2(u) - 1 du
____________ = eq. 11.8-41b
cosh2(u) √cosh2(u) - 1
|
1
16
|
|
du
=
cosh2(u)
|
1
16
|
|
(tanh2(u) - 1) du =
|
1
tanh(u) + C =
16
|
1
16
|
sinh(u)
+ C =
cosh(u)
|
1
16
|
____________
√cosh2(u) - 1
+ C
cosh(u)
|
You had to know in advance that
tanh(u)
was the integral of
tanh2(u) - 1,
but that's just the
analog of
tan(u) being the integral of
tan2(u) + 1, and you
probably already knew that one.
From here you should be able to finish the back substitution to x
and end up with equation 11.8-40.
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