1)
The example is to integrate
|
______
√1 - x2
dx eq. 11.8-27
x
|
Simple substitution of
u = 1 - x2 won't work on this one because
there's nothing you can do about the
dx.
Trig substitution of
sin(u) = x does work, but it is
just one of several successful attacks you can make on this one.
The
other kind of attack is to manipulate it into algebraic equivalents of the integrand that
we do know how to integrate using simple substitution. Two possible manipulations pop into my head
whenever I see a quotient like this. One is to multiply the numerator and
denominator by the numerator. The other is to multiply the numerator and
denominator by the denominator.
Either one of these attacks leads
to a solution. So I will show you both because on an exam you would have
flipped a coin in your head to choose one, and then followed the one you
chose to its conclusion. There's no real way to tell up front which is
the shorter road. Here is the longer road first:
Multiply numerator and denominator by the numerator. This gives
|
1 - x2
______ dx eq. 11.8-28a
x √1 - x2
|
Now you can break it up into two pieces
|
1 - x2
______ dx =
x √1 - x2
|
|
dx
______ -
x √1 - x2
|
|
x dx
______ eq. 11.8-28b
√1 - x2
|
The right-hand piece succumbs easily to
simple substitution.
The left-hand piece needs more manipulation. To get the
x by itself
in the numerator of that piece, we multiply its numerator and denominator by
x.
|
1 - x2
______ dx =
x √1 - x2
|
|
x dx
______ -
x2 √1 - x2
|
|
x dx
______ eq. 11.8-28c
√1 - x2
|
Now we can
substitute
u = 1 - x2
|
hence
|
1 - u = x2
|
and
|
1
- du = x dx eq. 11.8-28d
2
|
We get
|
1 - x2
______ dx =
x √1 - x2
|
1
−
2
|
|
du
_
(1 - u)√u
|
1
+
2
|
|
du
_ eq. 11.8-28e
√u
|
We could go and integrate the right-hand piece at this point. But
we still have another step to go before we integrate the left-hand piece.
So I'm going to wait. To integrate the left-hand piece we employ the
trick we discussed earlier about substituting for square
roots.
v2 = u
|
and
|
2v dv = du eq. 11.8-29a
|
Just for thrills, I'm going to apply this substitution to
both the
right-hand and left-hand
pieces.
1
-
2
|
|
du
_
(1 - u)√u
|
1
+
2
|
|
du
_ = -
√u
|
|
dv
+
1 - v2
|
|
dv eq. 11.8-29b
|
I'm going to stop here and do the second approach until it leads to the
identical situation as we have above:
Multiply numerator and denominator by the denominator.
This gives
|
______
x √1 - x2
dx eq. 11.8-30a
x2
|
This setup immediately lets us substitute
u = 1 - x2
|
hence
|
1 - u = x2
|
and
|
1
- du = x dx eq. 11.8-28d
2
|
to get
|
______
x √1 - x2
dx =
x2
|
1
−
2
|
|
_
√u du
eq. 11.8-30b
1 - u
|
Now apply the substitution
v2 = u
|
and
|
2v dv = du eq. 11.8-29a
|
and you get
1
−
2
|
|
_
√u du
= -
1 - u
|
|
v2 dv
eq. 11.8-30c
1 - v2
|
I'll leave it to you to apply
polynomial long division
to this and apply the
remainder rule to
the result to get back equation 11.8-29b
Or we could have used trig substitution.
The substitution is
sin(u) = x
|
and
|
___________
cos(u) du = √1 - sin2(u) du = dx eq. 11.8-31a
|
Pursuing this line of attack, we find
|
1 - sin2(u)
du =
sin(u)
|
|
cos2(u) sin(u)
du = eq. 11.8-31b
sin2(u)
|
|
cos2(u) sin(u)
du
1 - cos2(u)
|
Now substitute
v = cos(u)
|
and
|
dv = -sin(u) du eq. 11.8-31c
|
and we find ourselves right back at right-hand side of equation 11.8-30c,
which, in turn, leads back to equation 11.8-29b.
The point is that even with integrals that appear hard at first, there is often
more than one way that you can arrive at useful results.
To finish this one, we observe that the right-hand side of equation
11.8-29b consists of an integral that we have
done
several times before
by partial fractions
plus a second integral that is trivial. The result of all this is
-
|
|
dv
+
1 - v2
|
|
dv =
|
1
- ln
2
|
|
1 + v
1 - v
|
|
+ v + C eq. 11.8-32a
|
Now substitute back using equation
11.8-29a.
1
- ln
2
|
|
1 + v
1 - v
|
|
1
+ v + C = - ln
2
|
|
_
1 + √u
_
1 - √u
|
|
_
+ √u + C eq. 11.8-32b
|
Now substitute back again using equation
11.8-28d
1
- ln
2
|
|
_
1 + √u
_
1 - √u
|
|
_
+ √u + C = eq. 11.8-32c
|
1
- ln
2
|
|
______
1 + √1 - x2
______
1 - √1 - x2
|
|
______
+ √1 - x2 + C
|
We can simplify the stuff inside the log by multiplying both its
numerator and denominator by its numerator. We end up with
perfect squares on both top and bottom, so we can use
log identities to
carry the one-half into the log and to flip the result to
get rid of the minus sign to boot
(be sure to do all these steps yourself on paper). We end up with
|
______
√1 - x2
dx =
x
|
ln
|
|
x
______
1 + √1 - x2
|
|
______
+ √1 - x2 + C eq. 11.8-32d
|
Taking the derivative of this result is an interesting adventure, and you
should try it. It's easier if you break the log expression up into the
log of the numerator minus the log of the denominator. There are lots of
cancellations and simplifying to be done as you go, and with no mistakes
you will end up with the original integrand.
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