You can use * substitution* to convert a
complicated integral into a simpler one. In these problems, I'll let
u equal some convenient x-stuff --- say . To complete the
substitution, I must also substitute for . To do this, compute
, so . Then .

* Example.* Compute .

Here's what's going on. By the Chain Rule,

By the definition of antiderivative,

Now if , I have

So

The manipulations with and are just a convenient
way of doing the substitution. These are *not* the same
" " and " " we used in
discussing differentials.

* Example.* Compute .

* Example.* Later on, I'll derive the integration
formula

Use this formula to compute .

* Example.* Compute .

Notice that in the second step in the last example, the x's cancelled out, leaving only u's. If the x's had failed to cancel, I wouldn't have been able to complete the substitution.

But what made the x's cancel? It was the fact that I got an x from the derivative of . This leads to the following rule of thumb.

* Example.* Compute .

* Example.* Compute .

* Example.* Compute .

* Example.* Compute .

* Example.* Compute .

* Example.* Compute .

The next problem introduces a new idea. In some cases, to replace the
x's with u's, *you may need to solve the substitution equation
for* x.

* Example.* Compute .

There is no valid algebra which will allow me to multiply this out --- unless I plan to multiply out !

I'll let , so . If I stopped with that, I'd have

I can't continue as-is, because I have both x's and u's in the integral.

To get rid of the x's, I solve the substitution equation for x, to get . I can plug this into to get everything in terms of u. Here's the work:

* Example.* Compute .

In this problem, after making the substitution , I solve the substitution equation for x to get . Then I plug this into to get rid of the x's. Here's the work:

Copyright 2018 by Bruce Ikenaga