The problem is to use trigonometric substitution to find the indefinite integral:
|
dx
_______
√1 - 9x2
|
If you can get past the shock of seeing a slightly different form
from what appeared in the main text, this one is surprisingly easy.
Step 1: Make the substitution. You have
9x2 = (3x)2.
That makes the correct substitution be
sin(u) = 3x
|
and
|
du
cos(u) = 3
dx
|
or
|
1
cos(u) du = dx
3
|
Putting that in for
x and
dx gives
|
dx
_______ =
√1 - 9x2
|
1
3
|
|
cos(u) du
___________ =
√1 - sin2(u)
|
1
3
|
|
cos(u) du
=
cos(u)
|
1
3
|
|
du
|
Step 2: Integrate. That integral on the right is a piece of cake. You get
1
3
|
|
du =
|
u
+ C
3
|
Step 3: Substitute back. Your original substitution was
sin(u) = 3x, so you have to take
the inverse function to substitute back the bare
u. You will
end up with
|
dx
_______ =
√1 - 9x2
|
1
arcsin(3x) + C
3
|
If you did all of this shown here, good. That means you practiced
the trigonometric substitution technique. If, on the other hand,
you used the example in the main text, where we integrated
|
dx
______
√1 - x2
|
and applied equation
11.2-9 to that solution
as a short cut to the solution we arrived at here, that's good too. It means
you showed initiative and that you are keeping the material from previous
sections at your command. Finding shortcuts is one of the things mathematicians do,
so you demonstrated that you can think like a mathematician.
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