Cálculo de una primitiva.
Hemos de calcular las primitivas de $\mbox{sen}\, x\, e^{-i4n x}$ integrando por partes dos veces:
$$I=\int\mbox{sen}\, x\, e^{-i4n x}\, dx=\left\{\begin{array}{lll} u=\mbox{sen}\, x &,& du=\cos x\, dx \\
dv=e^{-i4n x}\, dx &,& v=\frac{-1}{4ni}e^{-i4n x}\end{array}\right\}= \frac{i}{4n}\left(e^{-i4n x}\mbox{sen}\, x-\int\cos x e^{-i4n x}\, dx\right)=$$
$$=\left\{\begin{array}{lll} u=\cos x &,& du=-\mbox{sen}\, x\, dx \\
dv=e^{-i4n x}\, dx &,& v=\frac{i}{4n}e^{-i4n x}\end{array}\right\}= \frac{i}{4n}\left(e^{-i4n x}\mbox{sen}\, x-\frac{i}{4n}e^{-i4n x}\cos x-\frac{i}{4n}\int\mbox{sen}\, x\, e^{-i4n x}\, dx\right)=$$
$$=\frac{1}{16n^2}\left(4ni\,\mbox{sen}\, x+\cos x\right)e^{-i4n x}+\frac{1}{16n^2}I$$
o bien
$$\left(1-\frac{1}{16n^2}\right)I=\frac{1}{16n^2}\left(4ni\,\mbox{sen}\, x+\cos x\right)e^{-i4n x}$$
de donde
$$I=\frac{1}{16n^2-1}\left(4ni\,\mbox{sen}\, x+\cos x\right)e^{-i4n x}+C$$