Trigonometric Integration Question

**craig** · January 20th 2011, 06:07 AM

Hi, just a quick question.

Doing a bit of Fourier Series and I've come across this integral, is there a best way to go about doing this?

$\frac{2}{\pi} \int_0^{\pi} \cos{(x)} \sin{(nx)} dx$ , where $n$ is an integer $0 \leq n \leq \infty$ .

I've attempted by using the product to sum formulas, but this just ends up with at least two sides of tricky algebra.

Has anyone come across this before, or have any helpful 'shortcuts'?

Thanks in advance for your help

**Liverpool** · January 20th 2011, 07:23 AM

What do you mean with trick algebra?
It is easy.
All what you need is : $sin(A)cos(B)=\dfrac{1}{2} \left( sin(A-B) + sin(A+B) \right)$

**craig** · January 20th 2011, 07:47 AM

Originally Posted by Liverpool

What do you mean with trick algebra?
It is easy.
All what you need is : $sin(A)cos(B)=\dfrac{1}{2} \left( sin(A-B) + sin(A+B) \right)$

Ohh sorry, meant tricky algebra.

I've sort of done it with the product to sum formulas, was just wondering if anyone know of a shorter way.

**Prove It** · January 20th 2011, 08:38 AM

Originally Posted by craig

Ohh sorry, meant tricky algebra.

I've sort of done it with the product to sum formulas, was just wondering if anyone know of a shorter way.

That IS the shortest way...

**craig** · January 20th 2011, 09:06 AM

Ahh thought as much, merci

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