Help with an Integral

[akleron]

Hello !
I've been trying to solve the following integral, the only idea came to my mind was integration by parts.
Unfortunately it didn't get me anywhere.
I'm adding both the problem and my attempt , would love to have some help !
Thanks.

(n and m are given as natural numbers, beyond my understanding why is that matter btw)

my attempt:

 

[skeeter]

natural numbers > 0

[MATH]f(x)=\sin(nx)\cos(mx)[/MATH]
[MATH]f(-x) = \sin(-nx)\cos(-mx) = -\sin(nx)\cos(mx) = -f(x) \implies f(x) \text{ is an odd function}[/MATH]
[MATH]\int_{-a}^a f(x) \, dx = 0[/MATH]

 

[akleron]

natural numbers > 0

[MATH]f(x)=\sin(nx)\cos(mx)[/MATH]
[MATH]f(-x) = \sin(-nx)\cos(-mx) = -\sin(nx)\cos(mx) = -f(x) \implies f(x) \text{ is an odd function}[/MATH]
[MATH]\int_{-a}^a f(x) \, dx = 0[/MATH]

Ohh got it now ! thanks

Would still love some help with the integral itself!

 

[skeeter]

try using the product to sum identity ...

[MATH]\sin{A}\cos{B} = \dfrac{1}{2}\left[\sin(A+B) + \sin(A-B) \right][/MATH]

 

[akleron]

Thanks A lot !

 

[Steven G]

You really need to include dx's at the end of your integrals!!

Whenever an integral is from -a to +a I would always spend a few seconds to see if the integrand is even or odd. If it is odd, as skeeter point out, then the answer is 0. If it even then you can just integrate from 0 to +a and then double your answer. Evaluating at x=0 is usually easier than evaluating at x=-a.

 

[akleron]

You really need to include dx's at the end of your integrals!!

Whenever an integral is from -a to +a I would always spend a few seconds to see if the integrand is even or odd. If it is odd, as skeeter point out, then the answer is 0. If it even then you can just integrate from 0 to +a and then double your answer. Evaluating at x=0 is usually easier than evaluating at x=-a.

Both points are Important, thanks !