∫cos∧2(ωx φ)sin(ωx φ)dx

∫sin^3(x)cos^2(x)dx=∫sin^2(x)cos^2(x)sin(x)dx=-∫sin^2(x)cos^2(x)dcos(x)=∫[cos^2(x)-1]cos^2(x)dcos(x)

sin^2(x)cos^4(x)=1/4*sin²2xcos²x=1/4*(1-cos4x)/2*(1+cos2x)/2=1/16*(1+cos2x-cos4x-cos2xcos4

2sinx*cosx=sin2x(cosx)^2-(sinx)^2=cos2x.由第二个又可以推出1-2(sinx)^2=cos2x2(cosx)^2-1=cos2x这个题之用这些就可以做出来了.自己

∫(sinx)^4/(cosx)^2dx=∫(1-(cosx)^2)^2/(cosx)^2dx=∫(1+(cosx)^4-2(cosx)^2)/(cosx)^2dx=∫1/(cosx)^2+(cosx

原式等于：∫[1-cos^2(x)]/cos^3(x)dx=∫dx/cos^3(x)-∫dx/cos(x)=(secxtanx+ln|secx+tanx|)/2-ln|secx+tanx|+C

∫[cos^3(x)]/[sin^2(x)]dx=积分:(cos^2x)/(sin^2x)dsinx=积分:(1-sin^2x)/sin^2x)dsinx=积分;1/sin^2xdsin^2x-积分1

=$4dx/sin^2x=4$csc^2xdx=-4cotx+c

据:cos2(x)=cosx^2-sinx^2=2cosx^2-1得:2sin^2(x)sin^2(φ)+2cos^2(x)cos^2(φ)-cos2(x)cos^2(φ)=2sin^2(x)sin^

f(x)=2sin(2wx-π/6)+λw=1/3+k/2(k=0,+_1.)w=5/6f(x)=2sin(5/3x-π/6)+λ(1)T=6π/5(2)f(x)=2sin(5/3x-π/6)-根号2

原式=(1/2)∫(sin2x)^2dx=(1/4)∫(1+cos4x)dx=(1/4)(x+sin4x/4)+C=x/4+sin4x/16+C

把一个sin(x)拿出来∫sin^3(x)cos^2(x)dx=-∫sin^2(x)cos^2(x)d(cos(x))=-∫(1-cos^2)cos^2(x)d(cos(x))=-∫cos^2-cos

求采纳.再问：图不太清楚但谢谢啦😊

∫(cos^3x/sin^2x)dx=∫[(1-(sinx)^2]/(sinx)^2dsinx=∫[1/(sinx)^2-1]dsinx=-1/sinx-sinx+C

令z=ωx+φ,dz=ωdx∫cos²(ωx+φ)sin(ωx+φ)dx=∫cos²zsinz*(1/ωdz)=(1/ω)∫cos²zd(-cosz)=(-1/ω)*(1

f(x)=(sinωx+cosωx））×(cosωx-sinωx）+cosωx2sinωx＋t=cos2ωx+sin2ωx＋t=根号2sin（2ωx+四分之π）+t根据相邻对称轴间的距离求出周期就能求

原式=-∫cos²xdcosx=-cos³x/3+C再问：第一步能讲一下为什么吗？再答：dcosx=-sinxdx采纳吧

∫cos2xdx/(sin^2xcos^2x)=4∫cos2xdx/(2sinxcosx)^2=4∫cos2xdx/(sin2x)^2=2∫cos2xd(2x)/(sin2x)^2=2∫d(sin2x

2cosx(sinx-cosx)+1=2sinxcosx-2cosx^2+1=sin2x+1-2cosx^2=sin2x-cos2x=√2sin(2x-π/4)

sinx=2cosx,sin^2x=4cos^2xsin^2x=4-4sin^2x,sin^2x=4/5(cosx+sinx)/(cosx-sinx)+sin^2x=(1+tanx)/(1-tanx)

原式=∫dx/(cos²x(1+4tan²x))=∫d(tanx)/(1+4tan²x)=1/2∫d(2tanx)/(1+(2tanx)²)=arctan(2t