Calculas question??????

Question

I can't figure out how to integrate this.

∫ sin(nx)cos(mx) dx from 0 to 2pi

also n and m are integers

Answer 1

you have to integrate by parts

∫ sin(nx)cos(mx) dx

u=sin(nx)                    dv=cos(mx) dx

du=ncos(nx) dx             v=(sin(mx))/m

∫ sin(nx)cos(mx) dx=(sin(nx))(sin(mx))/m -∫ ((sin(mx))/m)ncos(nx) dx

∫ sin(nx)cos(mx) dx=(sin(nx))(sin(mx))/m -(n/m)∫ (sin(mx)cos(nx) dx

now again integrate by parts

u=cos(nx)                    dv=sin(mx) dx

du=-nsin(nx) dx             v=(-cos(mx))/m

∫ sin(nx)cos(mx) dx=(sin(nx))(sin(mx))/m -(n/m)

[(cos(nx))((-cos(mx))/m)- ∫-nsin(nx) ((-cos(mx))/m)dx]

∫ sin(nx)cos(mx) dx=(sin(nx))(sin(mx))/m -(n/m)

[(cos(nx))((-cos(mx))/m)- (n/m) ∫sin(nx) (cos(mx))dx]

∫ sin(nx)cos(mx) dx=(sin(nx))(sin(mx))/m +

(n/m^2)(cos(nx))((-cos(mx)))+ (n^2/m^2) ∫sin(nx) (cos(mx))dx

∫ sin(nx)cos(mx) dx - (n^2/m^2) ∫sin(nx) (cos(mx))dx =

(sin(nx))(sin(mx))/m +(n/m^2)(cos(nx))(-cos(mx))

(1-(n^2/m^2))(∫ sin(nx)cos(mx) dx) =

(sin(nx))(sin(mx))/m +(n/m^2)(cos(nx))(-cos(mx))

∫ sin(nx)cos(mx dx =

((sin(nx))(sin(mx))/m +(n/m^2)(cos(nx))(-cos(mx)))/

((m^2-n^2)/m^2)

∫ sin(nx)cos(mx dx =

((msin(nx)sin(mx)/(m^2-n^2)) +(n/m^2-n^2)(cos(nx))(-cos(mx)))

at zero the answer is n/(m^2-n^2)

at 2pi the answer is also n/(m^2-n^2)

so the answer is zero