Fourier Series: B(n) = (1/4pi) int[0,2pi] (pi - x)^2 sin(nx) dx, 0 < x < 2pi

[R.K.4.7]

. . . . .\(\displaystyle \displaystyle B(n)\, =\, \dfrac{1}{\pi}\, \int_0^{2\pi}\, f(x)\, \sin(nx)\, dx\)

. . . . .\(\displaystyle \displaystyle f(x)\, =\, \left(\dfrac{\pi\, -\, x}{2}\right)^2,\, \mbox{ for }\, 0\, <\, x\, <\, 2\pi\)

. . . . .\(\displaystyle \displaystyle B(n)\, =\, \dfrac{1}{4\pi}\, \int_0^{2\pi}\, (\pi\, -\, x)^2\, \sin(nx)\, dx\)

So, after doing the integration, I get this:

. . . . .\(\displaystyle \displaystyle \left[\dfrac{\cos(nx)}{n}\right]_0^{2\pi}\)

Here, I get confused. What is the answer?

Also, please let me know what happens in the case of (Sin):

. . . . .\(\displaystyle \displaystyle \left[\dfrac{\sin(nx)}{n}\right]_0^{2\pi}\)

 

[Deleted member 4993]

. . . . .\(\displaystyle \displaystyle B(n)\, =\, \dfrac{1}{\pi}\, \int_0^{2\pi}\, f(x)\, \sin(nx)\, dx\)

. . . . .\(\displaystyle \displaystyle f(x)\, =\, \left(\dfrac{\pi\, -\, x}{2}\right)^2,\, \mbox{ for }\, 0\, <\, x\, <\, 2\pi\)

. . . . .\(\displaystyle \displaystyle B(n)\, =\, \dfrac{1}{4\pi}\, \int_0^{2\pi}\, (\pi\, -\, x)^2\, \sin(nx)\, dx\)

So, after doing the integration, I get this:

. . . . .\(\displaystyle \displaystyle \left[\dfrac{\cos(nx)}{n}\right]_0^{2\pi}\)

Here, I get confused. What is the answer?

Also, please let me know what happens in the case of (Sin):

. . . . .\(\displaystyle \displaystyle \left[\dfrac{\sin(nx)}{n}\right]_0^{2\pi}\)

Your integration is incorrect - you are forgetting (π-x)² part of the integrand.

 

[R.K.4.7]

Your integration is incorrect - you are forgetting (π-x)² part of the integrand.

But what is the solution for term cosnx/n ????

Sorry, i meant that this term is a small part of the complete answer. So im not able to solve just this small part.
Also, these type of parts come in this chapter a lot . So if you can thoroughly make me understand how this thing works that would be really great !