|Friday: Quiz 2
Next: Trig subst.
We use trig. identities and compute the integral directly as follows:
This always works for odd powers of
What about even
Realize that we should write
. The rest is straightforward.
This example illustrates a method for computing integrals of trig
functions that doesn't require knowing any trig identities at all or
any tricks. It is very tedious though. We compute
using complex exponentials
. We have
The answer looks totally different, but is in fact the same function.
Here are some more identities that we'll use in illustrating some tricks
Here we used the substitution
Also, with the substitution
Key trick: Write
Here's one that combines trig identities with the funnest
variant of integration by parts. Compute
Let's use integration by parts.
The above integral becomes
This is familiar. Solve for
. We get