Question

求积分：高手请进

∫cos（x/2）cos（nx）d（x）=
求解。尽量详细。。

Answer 1

原式=∫[cos(x/2+nx)+cos(x/2-nx)]/2 dx
=1/2*∫cos(x/2+nx)dx+1/2*∫cos(x/2-nx)dx
=sin(x/2+nx)/(1+2n)+sin(x/2-nx)/(1-2n)+C

Answer 2

原式=2 ∫cos(x/2)cos(nx) d(x/2)
=2 ∫cos(nx) d[sin(x/2)]
=(2/n) ∫cos(nx) d[sin(nx/2)]
=(2/n) ∫[1-2(sin(nx/2)^2)] d[sin(nx/2)] 关键在于将cos(nx)按照2倍角展开
=(2/n) {∫d[sin(x/2)]-2∫[sin(nx/2)]^2 d[sin(nx/2)]}
=(2/n) [sin(nx/2)]-(2/3)[sin(nx/2)^3]
=(1/n)sin(nx/2)-(4/3n)[sin(nx/2)^3]