急求∫(上限π/2,下限-π/2)(x^3+sin^2x)cos^2xdx=
1个回答
x^3cos^2xdx+sin^2xcos^2xdx
x^3cos^2x是奇函数所以它的定积分为0
算出得0
相关问题
求定积分∫上限π/2下限-π/2 (1+x)cosx/1+cos^2xdx
∫上限π/2下限0 sin^3 xdx
求定积分x²cos2xdx上限为π下限为0
求定积分上限0 下限π/2 xsin2xdx
求定积分 ∫arctan^3x/√(1+x^2) 积分上限π/2 下限-π/2
∫上限π/2,下限-π/2(cosx+x)dx=
求定积分∫sin(x+π/3),上限π,下限π/3.
若sin(x-π)=2cos(2π-x),求(sin(π-x)+5cos(2π-x))/(3cos(π-x)-sin(-
定积分上限π/2下限0sin^6x/(sin^6x+cos^6x)dx
设M=∫π2−π2[sinx1+x2cos4xdx,N=∫π/2−π2](sin3x+cos4x)dx,P=∫π2−π2