sin^8x在派到0的定积分

starprince2022-10-04 11:39:543条回答

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iikankan 共回答了22个问题 | 采纳率95.5%: ∫(sinx)^8dx
=-∫ (sinx)^7 dcosx
= -cosx (sinx)^7 + ∫ 7(cosx)^2(sinx)^6 dx
=-cosx (sinx)^7 +7∫ (1- (sinx)^2)(sinx)^6 dx
8∫(sinx)^8dx= -cosx (sinx)^7 + 7∫(sinx)^6dx
∫(sinx)^8dx
= (1/8) [-cosx (sinx)^7 + 7∫(sinx)^6dx]
= (1/8) {-cosx (sinx)^7 + (7/6)[-cosx (sinx)^5 + 5∫(sinx)^4dx]}
=(1/8){ -cosx (sinx)^7 + (7/6)(-cosx (sinx)^5 + (5/4)[-cosx(sinx)^3+3∫(sinx)^2dx]) }
=(1/8){ -cosx (sinx)^7 + (7/6)(-cosx (sinx)^5 + (5/4)[-cosx(sinx)^3+(3/2)[x-sin2x/2]) } + C; 1年前

yunji521 共回答了841个问题 | 采纳率: =亅(0,pi/2)+亅(pi/2),pi) 对第二个积分代换u=pi-x代入得: 亅(pi/2,pi)dx=-亅(pi/2,0)du=亅(0,pi/2)du,所以: 亅(0,pi)sin^8xdx=2亅(0,pi/2)sin^8xdx =2(7/8)(5/6)(3/4)(1/2)(pi/2) =35pi/128; 1年前

求证:cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x) 求证:cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x) 还有，csc^4a(1-cos^4a)-2cot^2a wanghaiting19871年前1: wong7208 共回答了18个问题 | 采纳率94.4%

cos^8x-sin^8x-cos2x
=(cos^4x+sin^4x)(cos^2x+sin^2x)(cos^2x-sin^2x)-cos2x
=(cos^4x+sin^4x)*1*cos2x-cos2x
=[(cos^2x+sin^2x)^2-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2]cos2x-cos2x
=[1-2(sinxcosx)^2-1]cos2x
=-(2sinxcosx)^2/2*cos2x
=-sin^2 2x/2*co2x
=-(1+cos4x)*cos2x/4
1/8(cos6x - cos2x)
=1/8*(-2)*sin(8x/2)sin(4x/2)
=-1/4sin4xsin2x
=-1/4*(2sin2xcos2x)sin2x
=-1/4*2(sin2x)^2cos2x
=-1/4(1+cos4x)cos2x
所以cos^8x-sin^8x-cos2x = 1/8(cos6x - cos2x)

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