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17.4 Integration by Substitution
We will now use the chain rule to find some antiderivatives. Let
be a real
valued function that is continuous on some interval
and differentiable on
the
interior of
, and let
be a function such that
has an antiderivative
on some interval
. We will suppose that
and
. It then follows that
is continuous on
and differentiable on
, and
![\begin{displaymath}
(F\circ g)'(t)=(F'\circ g)(t)g'(t)=(f\circ g)(t)g'(t)
\end{displaymath}](img4163.gif) |
(17.32) |
for all
in the interior of
; i.e.,
is an antiderivative for
on
. Thus
![\begin{displaymath}
\int f\Big( g(t)\Big)g'(t)dt=F \Big( g(t)\Big) \mbox{ where } F(u)=\int f(u)du.
\end{displaymath}](img4165.gif) |
(17.33) |
There is a standard ritual
for using (17.33) to find
when an antiderivative
can be found for
. We write:
Let
. Then
(or
), so
![\begin{displaymath}
\int f\Big( g(t)\Big)g'(t)dt=\int f(u)du=F(u)=F\Big( g(t)\Big).
\end{displaymath}](img4170.gif) |
(17.34) |
In the first equality of (17.34) we replace
by
and
by
, and in the last step we replace
by
. Since we have never
assigned
any meaning to ``
" or ``
",
we should think of (17.34)
just as a mnemonic device for remembering (17.33).
17.35
Example.
Find
![$\displaystyle { \int {{\sin (\sqrt x)} \over {\sqrt x}}dx}$](img4175.gif)
.
Let
. Then
, so
Suppose we want to find
. If we had a
in the
denominator, this would be a simple problem. (In fact we just considered this
problem in the previous example.) We will now discuss a method of introducing
the missing
.
Suppose
is a function on an interval
such that
is never zero on
the interior of
, and suppose that
is an inverse function for
. Then
for all
in the interior of
, so
We now apply the ritual (17.34):
Let
. Then
, so
If we can find an antiderivative
for
, then
We have shown that if
is an inverse function for
, then
![\begin{displaymath}
\int f\Big(g(x)\Big)dx=H\Big( g(x)\Big) \mbox{ where } H(u)=\int f(u)h'(u)du
\end{displaymath}](img4188.gif) |
(17.36) |
There is a ritual
associated with this result also. To find
:
Let
. Then
so
.
Hence
![\begin{displaymath}
\int f\Big( g(x)\Big)dx=\int f(u)h'(u)du=H(u)=H\Big( g(x)\Big).
\end{displaymath}](img4192.gif) |
(17.37) |
17.38
Example.
To find
![$\int\sin (\sqrt x)dx$](img4179.gif)
.
Let
. Then
so
.
Thus
We can now use integration by parts to find
![$\int u\sin (u)du$](img4196.gif)
.
Let
Then
Hence
17.39
Example.
To find
![$\displaystyle { \int {1\over {e^x+e^{-x}}}dx}$](img4200.gif)
.
Let
. Then
so
.
17.40
Example.
To find
![$\int t\sqrt{t+1}\;dt$](img4205.gif)
.
Let
. Then
so
.
Hence
17.41
Example.
To find
![$\displaystyle \int {{ (1-x)^{2 \over 5} \over x^{12\over 5}}dx}$](img4210.gif)
.
Let
![$\displaystyle {u = {1-x \over x} = {1\over x} - 1}$](img4212.gif)
. Then
![$\displaystyle {du = -{1\over x^2}dx}$](img4213.gif)
,
and
Thus
17.42
Exercise.
A
Find the following antiderivatives:
- a)
-
.
- b)
-
.
- c)
-
.
- d)
-
.
- e)
-
.
- f)
-
.
- g)
-
.
Next: 17.5 Trigonometric Substitution
Up: 17. Antidifferentiation Techniques
Previous: 17.3 Integration by Parts
  Index
Ray Mayer
2007-09-07