∫(上π/2下0)(cosx)^2(sinx)^4dx解法我用的方法好像非常繁琐就是不停地用二倍角公式打开,有没有什么简便算法呢?

Question

∫(上π/2下0)(cosx)^2(sinx)^4dx解法
我用的方法好像非常繁琐就是不停地用二倍角公式打开,有没有什么简便算法呢?

Answer 1

∫(0->π/2)(cosx)^2(sinx)^4dx
=∫(0->π/2)(cosx)^2(sinx)^2(sinx)^2dx
=1/4∫(0->π/2)(sin2x)^2(1-cos2x)/2 dx
=1/8∫(0->π/2)(sin2x)^2dx -1/8∫(0->π/2)(sin2x)^2cos2x dx
=1/16∫(0->π/2)(1-cos4x)dx -1/16∫(0->π/2)(sin2x)^2cos2x d2x
=1/16∫(0->π/2)dx+1/16∫(0->π/2)cos4xdx-1/16∫(0->π/2)(sin2x)^2dsin2x
=1/16 x|(0->π/2)+1/64∫(0->π/2)cos4xd4x-1/48 * (sin2x)^3 |(0->π/2)
=π/32 + 1/64 * sin4x |(0->π/2) -1/48 *(0-0)
=π/32+1/64*(0-0)
=π/32