Tricky Integral (Possibly Integration by Parts)

[npaul]

Hello, I'm new to this forum and pleased to be here!

My question is related to a rather tricky integral that I came across in my textbook. I can't solve it and it's really bothering me! I tried integration by parts and that was without success. I'd be very grateful for any help with this!

Here's the (updated) integral:
\(\displaystyle \int_{0}^{\frac{\pi }{2}}{\left( \cos ^{n}x \right)\left( \cos nx \right)dx}\)
where n is a nonnegative integer.

Thanks in advance,
Nico

 

[Deleted member 4993]

Hello, I'm new to this forum and pleased to be here!

My question is related to a rather tricky integral that I came across in my textbook. I can't solve it and it's really bothering me! I tried integration by parts and that was without success. I'd be very grateful for any help with this!

Here's the integral:

where n is a nonnegative integer.

Thanks in advance,
Nico

As written

cosn = constant

Then it is a standatrd integrant.

 

[npaul]

So sorry, I incorrectly copied the integral. This is what it actually is:
\(\displaystyle \int_{0}^{\frac{\pi }{2}}{\left( \cos ^{n}x \right)\left( \cos nx \right)dx}\)

Sorry about that, and thanks for the help anyways.
Nico

 

[royhaas]

The way to start this is by integrating by parts twice to develop a second order recursion relation. Note that the case of \(\displaystyle n=0,1\) are quite easy.