I know I am supposed to strip one layer away, in this case from sin, to give me:

But I don't know what to do next, only that it will probably involve u substitution.

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- September 8th 2008, 09:43 AMredman223[SOLVED] Using both u and trig substitution

I know I am supposed to strip one layer away, in this case from sin, to give me:

But I don't know what to do next, only that it will probably involve u substitution. - September 8th 2008, 09:48 AM11rdc11
On the first and then expand it out

On the second use half angle identity again - September 8th 2008, 09:53 AMMatt Westwood
You can solve this by parts, but it's tedious.

Pick either the sin or cos part to be u, and the other to be dv. You will end up with a bit you can evaluate and another bit you can integrate by parts again.

Then you integrate by parts again and you get another integration with reduced indices.

You will be able to deduce a rule which goes something like this (from my book of integrals):

or

- September 8th 2008, 10:02 AMredman223
- September 8th 2008, 10:22 AMTwistedOne151
Once to

Use u-substitution:

so that

Now expand, and pay attention to the change in the limits when doing the u-substitution

--Kevin C.

(Or if you knew about the beta and gamma functions, you could use the formula ) - September 8th 2008, 11:32 AMredman223

This is what I came up with so far.

What do you mean about the change in limits? - September 8th 2008, 11:39 AMChop Suey
- September 8th 2008, 11:40 AMMoo
It's x going from 0 to

If you change x ---> u=cos(x)

The boundaries of the integrals *have to* be the ones for u, not for x.

If x=0 ---> u=...

If x=pi/2 ---> u=...

These are the new boundaries/limits of the integral :)

So either you change it, either you substitute back u by cos x the integral you will get. But here, it'll be easier to change the boundaries/limits

NB : what you've got so far is correct - September 8th 2008, 11:41 AMredman223
Ok thanks, now I get it, I have to plug in the limits to cosx and re-evaluate the limits right? So it should be from 1 to 0 right?

- September 8th 2008, 11:44 AMMoo
- September 8th 2008, 11:44 AMChop Suey
- September 8th 2008, 12:07 PMredman223
I can't seem to come up with the right answer, but this is what I have so far:

- September 8th 2008, 12:08 PMMoo
- September 8th 2008, 12:09 PMShyam
- September 8th 2008, 01:42 PMChris L T521
See here to learn how to tackle integrals of this type.

--Chris