sint^4-sint^6的积分
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] - 作业帮
解歪18712856030…… [答案] 估计你的书本应该有这样一条公式: 当n为正整偶数时,即n=2m,m=1,2... ∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2) 当n为正整奇数时,即n=2m+1,m=0,1,2... ∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!] 知道这些后就好办了 ∫(0→π/2)(sinx)^4dx =(3/4)*(1/2)*(π/2...
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] dt - 作业帮
解歪18712856030…… [答案] 这里用一个公式会简单些:∫ [0--->π/2] f(sinx)dx=∫ [0--->π/2] f(cosx)dx∫[0→π/2] (sin⁴t-sin⁶t) dt=∫[0→π/2] sin⁴t(1-sin²t) dt=∫[0→π/2] sin⁴tcos²t dt=1/2( ...
@阴胞419：∫(上限π/2 下限0) [(sint)^4 - (sint)^6] dt - 作业帮
解歪18712856030…… [答案] 当n为正整偶数时,即n=2m,m=1,2...∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2)当n为正整奇数时,即n=2m+1,m=0,1,2...∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!]∫(0→π/2)(sinx)^4dx=(3/4)*(1/2)*(π/2)=3π/16∫(0...
@阴胞419：(sin^4 - sin^6)dt的不定积分怎么求 ? -
解歪18712856030…… (sin^4-sin^6)dt的不定积分是x/16-(1/64)cos4x-(1/48)(sin2x)^3+C. 解: ∫[(sinx)^4-(sinx)^6]dt=∫(sinx)^4[1-(sinx)^2]dt=∫(sinx)^4(cosx)^2dt=(1/4)∫(2sinxcosx)^2(sinx)^2dt=(1/8)∫(sin2x)^2[2(sinx)^2]dt=(1/8)∫(sin2x)^2(1-cos2x)dt=(1/8)∫(sin2x)^2dx-(1/8)...
@阴胞419：sinx4 - sinx6的定积分 - 作业帮
解歪18712856030…… [答案] ∫ [(sinx)^4- (sinx)^6] dx=(1/4)∫ (1- cos2x)^2 dx - (1/8)∫ (1-cos2x)^3 dx=(1/4)∫ [1- 2cos2x + (cos2x)^2] dx - (1/8)∫ [1-3cos2x +3(cos2x)^2+ (cos2x)^3] dx=(1/8)∫ [3- 4cos2x -cos4x] dx - (1/16)∫ [...
@阴胞419：(sinx)^4 - (sinx)^6的积分 -
解歪18712856030…… 如果是不定积分需要用倍角公式先降幂,降成一次幂后积分便可.如果是定积分,则有公式直接计算可得.
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] dt求过程啊...谢谢啦 -
解歪18712856030…… 估计你的书本应该有这样一条公式: 当n为正整偶数时,即n=2m,m=1,2... ∫(0→π/2)(sinx)^ndx=[(2m-1)!!/(2m)!!](π/2) 当n为正整奇数时,即n=2m+1,m=0,1,2... ∫(0→π/2)(sinx)^ndx=[(2m)!!/(2m+1)!!] 知道这些后就好办了 ∫(0→π/2)(sinx)^4dx =(3/4)*(1/2)*(π/2) =3π/16 ∫(0→π/2)(sinx)^6dx =(5/6)*(3/4)*(1/2)*(π/2) =5π/32 所以,原式=3π/16-5π/32=π/32
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] dt -
解歪18712856030…… 这里用一个公式会简单些:∫ [0--->π/2] f(sinx)dx=∫ [0--->π/2] f(cosx)dx ∫[0→π/2] (sin⁴t-sin⁶t) dt=∫[0→π/2] sin⁴t(1-sin²t) dt=∫[0→π/2] sin⁴tcos²t dt=1/2( ∫[0→π/2] sin⁴tcos²t dt+∫[0→π/2] sin²tcos⁴t dt )=1/2 ∫[0→π/2] sin²tcos²t(sin²t+cos...
@阴胞419：(sinx)^4 - (sinx)^6的积分 - 作业帮
解歪18712856030…… [答案] ∫(sinx)^ndx=(-1/n)(sinx)^(n-1)·cosx+[(n-1)/n]∫(sinx)^(n-2)dx ∴ ∫(sinx)^6dx=(-1/6)(sinx)^5·cosx+(5/6)∫(sinx)^4dx 故原式=(1/6)(sinx)^5·cosx-(1/24)sin³xcosx-(1/16)sinxcosx+(1/16)x+C
@阴胞419：∫(上限π/2 下限0) [(sint)^4 - (sint)^6] dt -
解歪18712856030…… 当n为正整偶数时,即n=2m,m=1,2... ∫(0→π/2)(sinx)^ndx=[(2m-1)!!/(2m)!!](π/2) 当n为正整奇数时,即n=2m+1,m=0,1,2... ∫(0→π/2)(sinx)^ndx=[(2m)!!/(2m+1)!!] ∫(0→π/2)(sinx)^4dx =(3/4)*(1/2)*(π/2) =3π/16 ∫(0→π/2)(sinx)^6dx =(5/6)*(3/4)*(1/2)*(π/2) =5π/32 3π/16-5π/32=π/32
解歪18712856030…… [答案] 估计你的书本应该有这样一条公式: 当n为正整偶数时,即n=2m,m=1,2... ∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2) 当n为正整奇数时,即n=2m+1,m=0,1,2... ∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!] 知道这些后就好办了 ∫(0→π/2)(sinx)^4dx =(3/4)*(1/2)*(π/2...
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] dt - 作业帮
解歪18712856030…… [答案] 这里用一个公式会简单些:∫ [0--->π/2] f(sinx)dx=∫ [0--->π/2] f(cosx)dx∫[0→π/2] (sin⁴t-sin⁶t) dt=∫[0→π/2] sin⁴t(1-sin²t) dt=∫[0→π/2] sin⁴tcos²t dt=1/2( ...
@阴胞419：∫(上限π/2 下限0) [(sint)^4 - (sint)^6] dt - 作业帮
解歪18712856030…… [答案] 当n为正整偶数时,即n=2m,m=1,2...∫(0→π/2)(sinx)^ndx=[(2m-1)!/(2m)!](π/2)当n为正整奇数时,即n=2m+1,m=0,1,2...∫(0→π/2)(sinx)^ndx=[(2m)!/(2m+1)!]∫(0→π/2)(sinx)^4dx=(3/4)*(1/2)*(π/2)=3π/16∫(0...
@阴胞419：(sin^4 - sin^6)dt的不定积分怎么求 ? -
解歪18712856030…… (sin^4-sin^6)dt的不定积分是x/16-(1/64)cos4x-(1/48)(sin2x)^3+C. 解: ∫[(sinx)^4-(sinx)^6]dt=∫(sinx)^4[1-(sinx)^2]dt=∫(sinx)^4(cosx)^2dt=(1/4)∫(2sinxcosx)^2(sinx)^2dt=(1/8)∫(sin2x)^2[2(sinx)^2]dt=(1/8)∫(sin2x)^2(1-cos2x)dt=(1/8)∫(sin2x)^2dx-(1/8)...
@阴胞419：sinx4 - sinx6的定积分 - 作业帮
解歪18712856030…… [答案] ∫ [(sinx)^4- (sinx)^6] dx=(1/4)∫ (1- cos2x)^2 dx - (1/8)∫ (1-cos2x)^3 dx=(1/4)∫ [1- 2cos2x + (cos2x)^2] dx - (1/8)∫ [1-3cos2x +3(cos2x)^2+ (cos2x)^3] dx=(1/8)∫ [3- 4cos2x -cos4x] dx - (1/16)∫ [...
@阴胞419：(sinx)^4 - (sinx)^6的积分 -
解歪18712856030…… 如果是不定积分需要用倍角公式先降幂,降成一次幂后积分便可.如果是定积分,则有公式直接计算可得.
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] dt求过程啊...谢谢啦 -
解歪18712856030…… 估计你的书本应该有这样一条公式: 当n为正整偶数时,即n=2m,m=1,2... ∫(0→π/2)(sinx)^ndx=[(2m-1)!!/(2m)!!](π/2) 当n为正整奇数时,即n=2m+1,m=0,1,2... ∫(0→π/2)(sinx)^ndx=[(2m)!!/(2m+1)!!] 知道这些后就好办了 ∫(0→π/2)(sinx)^4dx =(3/4)*(1/2)*(π/2) =3π/16 ∫(0→π/2)(sinx)^6dx =(5/6)*(3/4)*(1/2)*(π/2) =5π/32 所以,原式=3π/16-5π/32=π/32
@阴胞419：∫(0→π/2) [(sint)^4 - (sint)^6] dt -
解歪18712856030…… 这里用一个公式会简单些:∫ [0--->π/2] f(sinx)dx=∫ [0--->π/2] f(cosx)dx ∫[0→π/2] (sin⁴t-sin⁶t) dt=∫[0→π/2] sin⁴t(1-sin²t) dt=∫[0→π/2] sin⁴tcos²t dt=1/2( ∫[0→π/2] sin⁴tcos²t dt+∫[0→π/2] sin²tcos⁴t dt )=1/2 ∫[0→π/2] sin²tcos²t(sin²t+cos...
@阴胞419：(sinx)^4 - (sinx)^6的积分 - 作业帮
解歪18712856030…… [答案] ∫(sinx)^ndx=(-1/n)(sinx)^(n-1)·cosx+[(n-1)/n]∫(sinx)^(n-2)dx ∴ ∫(sinx)^6dx=(-1/6)(sinx)^5·cosx+(5/6)∫(sinx)^4dx 故原式=(1/6)(sinx)^5·cosx-(1/24)sin³xcosx-(1/16)sinxcosx+(1/16)x+C
@阴胞419：∫(上限π/2 下限0) [(sint)^4 - (sint)^6] dt -
解歪18712856030…… 当n为正整偶数时,即n=2m,m=1,2... ∫(0→π/2)(sinx)^ndx=[(2m-1)!!/(2m)!!](π/2) 当n为正整奇数时,即n=2m+1,m=0,1,2... ∫(0→π/2)(sinx)^ndx=[(2m)!!/(2m+1)!!] ∫(0→π/2)(sinx)^4dx =(3/4)*(1/2)*(π/2) =3π/16 ∫(0→π/2)(sinx)^6dx =(5/6)*(3/4)*(1/2)*(π/2) =5π/32 3π/16-5π/32=π/32
