∫［π/2,0］sinx^4dx

∫(0,π/2)cos(sinx)dx∫(0,π/2)cos(sinx)dx∫(0,π/2)cos(sinx)dx=sin(cos(π/2-0)dx

∫(2π,0)|sinx|dx=∫(2π,0)|sinx|dx=∫(2π,0)|sinx|dx=∫（2π,0）|sinx|dx=∫（π,0）sinxdx+∫（2π,π）（-sinx）dx=2+2=4如

∫0~2πx|sinx|dx∫0~2πx|sinx|dx∫0~2πx|sinx|dx∵当0

比较∫sin(sinx)dx与∫cos(sinx)dx在(0,π/4)大小比较∫sin(sinx)dx与∫cos(sinx)dx在(0,π/4)大小比较∫sin(sinx)dx与∫cos(sinx)d

∫(0,10π)[(sinx)^3]/[2(sinx)^2+(cosx)^4]dx∫(0,10π)[(sinx)^3]/[2(sinx)^2+(cosx)^4]dx∫(0,10π)[(sinx)^3]

∫x.√(sinx^2-sinx^4）dx（下限0上限π）∫x.√(sinx^2-sinx^4）dx（下限0上限π）∫x.√(sinx^2-sinx^4）dx（下限0上限π）∫(0→π)√(sin&#

∫［0,2π］|sinx|dx求定积分∫［0,2π］|sinx|dx求定积分∫［0,2π］|sinx|dx求定积分∫［0,2π］|sinx|dx=∫［0,π］sinxdx-∫［π,2π］sinxdx=

∫(sinx)^2dx∫(sinx)^2dx∫(sinx)^2dx∫(sinx)^2dx=1/2∫(1-cos2x)dx=x/2-sin2x/4+C

为什么∫(2π-0)|sinx|dx=∫（0,π）sinxdx-∫（π,2π）sinxdx为什么∫(2π-0)|sinx|dx=∫（0,π）sinxdx-∫（π,2π）sinxdx为什么∫(2π-0)

求积分∫(0→2π)f（sinx^2）*sinx^3dx∫(0→2π)f（sinx^2）*sinx^3dx求积分∫(0→2π)f（sinx^2）*sinx^3dx∫(0→2π)f（sinx^2）*si

∫x/［（sinx）＾2］dX＝∫x/［（sinx）＾2］dX＝∫x/［（sinx）＾2］dX＝

∫(0~π)sinx/(5-4cosx)dx怎么求?∫(0~π)sinx/(5-4cosx)dx怎么求?∫(0~π)sinx/(5-4cosx)dx怎么求?∫(0~π)sinx/(5-4cosx)dx

∫π/2→0(2x+sinx)dx=?∫π/2→0(2x+sinx)dx=?∫π/2→0(2x+sinx)dx=?上限是π/2,下限是0么那么∫(2x+sinx)dx=x²-cosx（代入上

∫(0~π/2）(sinx)^3dx=?∫(0~π/2）(sinx)^3dx=?∫(0~π/2）(sinx)^3dx=?∫(sinx)^3dx=∫(sin²x)sinxdx=-∫(1-cos

∫(下限0,上限2π)|sinx|dx∫(下限0,上限2π)|sinx|dx∫(下限0,上限2π)|sinx|dx∫[0,2π]|sinx|dx=∫[0,π]sinxdx+∫[π.2π]-sinxdx

∫(0,π)|sinx-cosx|dx∫(0,π)|sinx-cosx|dx∫(0,π)|sinx-cosx|dx∫(0,π)|sinx-cosx|dx=∫(0,π)√2|sin(x-π/4)|dx=

∫(4,0)(x^2+1+sinx)dx∫(4,0)(x^2+1+sinx)dx∫(4,0)(x^2+1+sinx)dx∫(4,0)(x^2+1+sinx)dx=(1/3x^3+x-cosx)|(4,

∫(sinx)^4×(cosx)^2dx∫(sinx)^4×(cosx)^2dx∫(sinx)^4×(cosx)^2dx∫(sinx)^4*(cosx)^2*dx(线分部积分)=-1/3*(cosx)

∫(sinx)^4/(cosx)^2dx过程∫(sinx)^4/(cosx)^2dx过程∫(sinx)^4/(cosx)^2dx过程显然1-sin²x=cos²x那么∫(sinx)

如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx如何证明∫[0,π]xf(sinx)dx=π∫[0,π/2]f(sinx)dx如何证明∫[0,π]xf(sinx)dx=π