∫x^3*sin^2x (x^4 2x^2 1)dx (上限为5 下限为-5)

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∫x^3*sin^2x (x^4 2x^2 1)dx (上限为5 下限为-5)
求微积分 ∫sin^2(x)cos^4(x) dx

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

化简2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1

2sin^2[(π/4)+x]+根号3(sin^x-cos^x)-1=-(1-2sin^2[(π/4)+x)-√3cos2x=-cos(π/2+2x)-√3cos2x=sin2x-√3cos2x=2[

三角等式求证:cos^6x+sin^6x=1-3sin^2x+3sin^4x

用公式a³+b³=(a+b)(a²-ab+b²)cos^6x+sin^6x=(cos²x)³+(sin²x)³=(cos

∫ 上√3下-√3 (x^2 sin x)/(1+x^4) dx

0,到根号3,与-根号3到0的积分互为相反数,加起来=0

∫ sin^4(x)cos^3(x)dx

∫sin^4(x)cos^3(x)dx=∫sin^4(x)cos^2(x)cosxdx=∫sin^4(x)*[1-sin^2(x)]d(sinx)=∫sin^4(x)d(sinx)-∫sin^6(x)

∫[1/(sin^2(x)cos^4(x)]dx

求采纳.再问:图不太清楚但谢谢啦😊

∫sin(x) cos^2(x)dx

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

求导f(x) = cos(3x) * cos(2x) + sin(3x) * sin(2x).

f(x)=cos(3x)*cos(2x)+sin(3x)*sin(2x)=cos(3x-2x)=cosxf'(x)=-sinx

求∫e^(3x)sin(4x)dx

I=1/3∫sin(4x)d(e^(3x))=1/3(sin(4x)e^(3x)-∫e^(3x)dsin(4x))=1/3sin(4x)e^(3x)-4/9∫cos(4x)de^(3x)=1/3sin

求极限 ((sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1

由和差化积公式分子=2sin[(x^3+x^2)/2]cos[(x^3+x^2-2x)/2]x→0,则(x^3+x^2)/2→0,sin则(x^3+x^2)/2和(x^3+x^2)/2是等价无穷小而c

s = 2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x)

x=0:0.1:2*pi;s=2*sin(x)-sin(2*x)+2/3*sin(3*x)-1/2*sin(4*x)+2/5*sin(5*x);plot(x,s)

已知函数f(x)=sin^2 x+2根号3sinxcosx+sin(x+π/4)sin(x-π/4),x属于R,求f(x

f(x)=sin^2x+2√3sinxcosx+sin(x+π/4)sin(x-π/4)=(1-cos2x)/2+√3sin2x+(1/2)2sin(x-π/4)cos(x-π/4)=2-2cos2x

化简[1-(sin^4x-sin^2cos^2x+cos^4x)/(sin^2)]+3sin^2x

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x

用换元法求下列积分∫x^3sin^x/(x^4+1)dx∫x^3sin^2x/(x^4+1)dx

换元法?没必要啊显然这是个奇函数而积分限关于原点对称所以原式=0

sin(5x)-sin(3x)= 根号2cos(4x)

1、sinx=2分之根号2,或者cos(4x)=02、sinx=负二分之一,或者cosx=二分之一,或者cosx=0

证明sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=3/4

sin^2(x)+cos^2(x+30)+sin(x)cos(x+30)=sin^2(x)+cos(x+30)[cos(x+30)+sinx]=sin^2(x)+cos(x+30)(cosxcos30

使用微积分基本原理 (1)∫(x sin√(x^2+4))/√(x^2+4) dx(2)∫x^2 sin(x^3+5)

(1)原式=1/2∫sin√(x²+4)/(√(x²+4)d(x²+4)=-∫sin√(x²+4)d√(x²+4)=cos√(x²+4)+C

∫dx/cos^2X+4sin^2X

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

(1-(sin^4x-sin^2xcos^2x+cos^4x)/sin^2x +3sin^2x

sin^4x-sin^2xcos^2x+cos^4x=sin^4x+2sin^2xcos^2x+cos^4x-3sin^2xcos^2x=(sin^2x+cos^2x)^2-3sin^2xcos^2x