4)
The example is to integrate
|
______
√tan(x) dx
|
As the lead-in said, we make the substitution for
tan(x).
u = tan(x)
|
and
|
du
= dx eq. 11.8-43a
u2 + 1
|
We get
|
______
√tan(x) dx =
|
|
_
√u du
eq. 11.8-43b
u2 + 1
|
Now we substitute for the square root as you learned to do in the main text.
v2 = u
|
and
|
2 v dv = du eq. 11.8-44a
|
and the integral becomes
|
_
√u du
= 2
u2 + 1
|
|
v2 dv
eq. 11.8-44b
v4 + 1
|
Recall that the method of
partial fractions
is applicable to
all functions that are the quotient of two polynomials,
and this one is just that. But recall also that the method of partial fractions
requires that you first
factor the denominator. This denominator is not
as hard to factor as you might imagine. You just have to take it step by step.
We assume that we will be able to factor it into two quadratics.
v4 + 1 = (v2 + pv + q)(v2 + rv + s) = eq. 11.8-45a
v4 + (p + r)v3 + (pr + q + s)v2 + (ps + rq)v + qs
Now we go and deduce what the coefficients,
p,
q,
r, and
s must
be equal to.
Since v4 + 1 has no cubed term, it must
be that p + r = 0, or equivalently
r = -p. So we make that substitution,
which eliminates the cube term from the equation (and we cancel
the v4 from both sides).
1 = (-p2 + q + s)v2 + p(s - q)v + qs eq. 11.8-45b
Since the
v term must also be zero, we conclude that
s = q. Making that substitution
eliminates the
v term.
1 = (-p2 + 2q)v2 + q2 eq. 11.8-45c
Clearly
q2 = 1, so
q = ±1. We know too that
the
v2 term must also be zero, and that makes it clear
that
-p2 + 2q = 0. That only
works if we choose
q = +1. We find that
p = ±√2.
And that finishes the factoring because now we know what all the coefficients
of the two quadratics are:
v4 + 1 = (v2 - √2v + 1)(v2 + √2v + 1) eq. 11.8-45d
If you don't believe me, multiply it out for yourself.
When we try to apply the quadratic formula
to either of these factors, we find that neither of them can be factored any further.
So here is the partial fraction problem we have to solve:
2
|
|
v2 dv
=
v4 + 1
|
|
(Av + B) dv
_ +
v2 - √2v + 1
|
|
(Cv + D) dv
_ eq. 11.8-46
v2 + √2v + 1
|
Heaviside can't help us with this one. We have to use the standard method. To
do that we have to multiply
(Av + B) by
(v2 + √2v + 1),
and
(Cv + D) by
(v2 - √2v + 1),
then isolate terms for each power of
v to form our four linear equations.
Matching those terms up with the numerator that's to the left of the equal, we have
(A + C)v3 = 0 eq. 11.8-47a
_ _
(B + D + √2A - √2C)v2 = 2v2
_ _
(A + C + √2B - √2D)v = 0
(B + D) = 0
As simultaneous linear equations go, these are pretty easy. We can see immediately
from the first equation that
C = -A and from
fourth equation that
D = -B, which eliminates
the first and fourth equations.
_
2√2Av2 = 2v2 eq. 11.8-47b
_
2√2Bv = 0
|
Clearly from this we get
_
√2
A =
2
|
and
|
_
√2
C = −
2
|
and
|
B = D = 0
|
So the partial fraction problem is now solved.
2
|
|
v2 dv
=
v4 + 1
|
_
√2
2
|
|
v dv
_ -
v2 - √2v + 1
|
_
√2
2
|
|
v dv
_ eq. 11.8-48
v2 + √2v + 1
|
The thrill of the chase is not over yet, though. We still have to
complete the square on each of these
partial fractions. We can actually do both of them at the same time
using the "
±" notation, where the "
+" refers to
the second partial fraction shown above and the "
-" refers
to the first one. When you see the "
" symbol,
it means just the opposite.
w = v ±
|
√2
2
|
and
|
_
w2 = v2 ± √2v +
|
1
eq. 11.8-49a
2
|
|
hence
|
w
|
√2
= v
2
|
and
|
w2 +
|
1
=
2
|
_
v2 ± √2v + 1
|
as well as
dw = dv.
Keeping in mind the sign notation and that we are proceeding from
here solving both partial fractions at once, the equation is
_
√2
2
|
|
dv
_ =
v2 ± √2v + 2
|
_
√2
2
|
|
w dw
+
w2 + 1/2
|
1
2
|
|
dw
eq. 11.8-49b
w2 + 1/2
|
Notice that
(w a) = w + a.
That's what accounts for the arrangement of signs you see in equation 11.8-49b.
The first summand yields to
simple substitution.
z = w2 +
|
1
2
|
and
|
1
dz = w dw eq. 11.8-50a
2
|
For the second summand,
what you've learned of
trig substitution tells
you to divide out the constant (I'll let you work the details of doing that)
and then substitute
_
tan(t) = √2 w
|
and
|
_
√2
2
|
(tan2(t) + 1) dt = dw eq. 11.8-50b
|
The substituted integrals end up being
_
√2
4
|
|
dz
+
z
|
_
√2
2
|
|
_
√2
dt = ln|z| +
4
|
_
√2
t + K eq. 11.8-50c
2
|
Now comes the arduous journey of back substitution from
z and
t to
w to
v to
u to
x.
_
√2
ln|z| +
4
|
_
√2
t + K = eq. 11.8-51a
2
|
_
√2
ln
4
|
|
w2 +
|
1
2
|
|
_
√2 _
+ arctan(√2w) + K =
2
|
_
√2
ln
4
|
|v2 ±
|
_
√2v + 1|
|
_
√2 _
+ arctan(√2v ± 1) + K =
2
|
_
√2 __
ln|u ± √2u + 1|
4
|
_
√2 __
+ arctan(√2u ± 1) + K =
2
|
_
√2 _______
ln|tan(x) ± √2tan(x) + 1|
4
|
_
√2 _______
+ arctan(√2tan(x) ± 1) + K
2
|
Remember that the above is an encapsulation of both partial fractions. So,
accounting for all the plusses and minuses and all the terms arising from each of the
partial fractions, the final answer is
|
______
√tan(x) dx =
|
_
√2 _ ______
arctan(√2√tan(x) + 1) + eq. 11.8-51b
2
|
_
√2 _ ______
arctan(√2√tan(x) - 1) +
2
|
_
√2 _ ______
ln|tan(x) - √2√tan(x) + 1| -
4
|
_
√2 _ ______
ln|tan(x) + √2√tan(x) + 1| + K
4
|
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