=∫(π/2,-π/2) √[cos^2 x(1-cos^2 x)] dx
=∫(π/2,-π/2) √[cos^2 x *sin^2 x] dx
=∫(π/2,-π/2) cosx*sinx dx
=∫(π/2,-π/2) sinx d(sinx)
= (1/2)(sinx)^2 |(π/2,-π/2)
= (1/2)(1-1)
=0
求定积分∫(π\/2,-π\/2) 根号cos^x-cos^4x dx
=∫(π\/2,-π\/2) √[cos^2 x *sin^2 x] dx =∫(π\/2,-π\/2) cosx*sinx dx =∫(π\/2,-π\/2) sinx d(sinx)= (1\/2)(sinx)^2 |(π\/2,-π\/2)= (1\/2)(1-1)=0
0到派x根号下cos^2x-cos^4xdx
=∫(π\/2,-π\/2) √[cos^2 x(1-cos^2 x)] dx =∫(π\/2,-π\/2) √[cos^2 x *sin^2 x] dx =∫(π\/2,-π\/2) cosx*sinx dx =∫(π\/2,-π\/2) sinx d(sinx)= (1\/2)(sinx)^2 |(π\/2,-π\/2)= (1\/2)(1-1)=0 ...
计算定积分∫(π\/2 ~ -π\/2) x(sinx+cosx)^2 dx 函数y=sin^4x+cos^4x...
=1\/2*x²-1\/2*xcos2x+1\/2*∫cos2x dx =1\/2*x²-1\/2*xcos2x+1\/4*∫cos2x d(2x)=1\/2*x²-1\/2*xcos2x+1\/4*sin2x 然后将积分上下限带进去计算即可 y=sin^4x+cos^4x y'=4sin³x*(sinx)'+4cos³x*(cosx)'=4sin³x*cosx-4cos³...
定积分∫(-π\/2,π\/2)(cos^4x+sin^3x)dx=
解：∵(cosx)^4是偶函数，(sinx)^3是奇函数 ∴∫<-π\/2,π\/2>(cosx)^4dx=2∫<0,π\/2>(cosx)^4dx ∫<-π\/2,π\/2>(sinx)^3dx=0 故 ∫<-π\/2,π\/2>((cosx)^4+(sinx)^3)dx =∫<-π\/2,π\/2>(cosx)^4dx+∫<-π\/2,π\/2>(sinx)^3dx =2∫<0,π\/2>(cosx)^4...
求指导,求∫[π\/2 -π\/2]sin^4xdx的定积分
2x) +cos(4x)\/8 ∫[π\/2 -π／2]sin⁴xdx =∫[π\/2 -π／2][9\/8-cos(2x) +cos(4x)\/8]dx =9x\/8 -sin(2x)\/2 +sin(4x)\/32|(π\/2，-π\/2) 积分上下限很不清楚，按你写的，则-π\/2是积分上限，就不再写了。你直接按上下限代进去就可以了。
...∫(上1下-1) xdx\/√(5-4x) 2.∫(上π\/2,下-π\/2) √(cosx-cos^3*x...
根据倍角公式1+cosx=2(cos(x\/2))^2 ∫ (π\/3→－π\/3)【(cos x)\/(1+cosx)】dx= ∫ (π\/3→－π\/3)【1-（1\/(1+cosx)）】dx= ∫ (π\/3→－π\/3)【1-1\/(2(cos(x\/2)^2)】dx=x-tan（x\/2）至此直接代入积分限就可以了 ...
求定积分,求解
回答：∫<0,π\/2>(cosx)^2(sinx)^2dx =(1\/4)∫<0,π\/2>(sin2x)^2dx =(1\/8)∫<0,π\/2>(1-cos4x)dx =(1\/8)(x-sin4x\/4)|<0,π\/2> =π\/16.
...∫(上1下-1) xdx\/√(5-4x) 2.∫(上π\/2,下-π\/2) √(cosx-cos^3*x...
http:\/\/hi.baidu.com\/fjzntlb\/album\/item\/2fba067b72b9dd8c2f73b377.html# 第二个没必要换元，实在要换就令t=cosx
求不定积分∫(cos^4x)dx,
先降次把cos^4x降为cos^2x*cos^2x再把cos^2x降为1\/2(cos2x+1)由于有两项这个式子相乘次数又升高了再次用倍角公式降次降到一次为止别忘了c
求定积分∫上限π\/2,下限0 4sin^2xcos^2xdx,过程?
定积分=1\/2∫sin^2(2x)d2x=1\/8∫(1-cos4x)d4x=(2π-1)\/8 ,,,（上下限不计）
