急!谁能求sin^4(x)的原函数?要过程.已知:F'(x)=sin^4(x)求:F(x)能否解释一下为什么:∫0→π/2:sin^4(x)=(3/4)*(1/2)*(1/2)π∫0→π/2:sin^6(x)=(5/6)*(3/4)*(1/2)*(1/2)π
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![急!谁能求sin^4(x)的原函数?要过程.已知:F'(x)=sin^4(x)求:F(x)能否解释一下为什么:∫0→π/2:sin^4(x)=(3/4)*(1/2)*(1/2)π∫0→π/2:sin^6(x)=(5/6)*(3/4)*(1/2)*(1/2)π](/uploads/image/z/12378430-46-0.jpg?t=%E6%80%A5%21%E8%B0%81%E8%83%BD%E6%B1%82sin%5E4%EF%BC%88x%EF%BC%89%E7%9A%84%E5%8E%9F%E5%87%BD%E6%95%B0%3F%E8%A6%81%E8%BF%87%E7%A8%8B.%E5%B7%B2%E7%9F%A5%EF%BC%9AF%27%28x%29%3Dsin%5E4%28x%29%E6%B1%82%EF%BC%9AF%28x%29%E8%83%BD%E5%90%A6%E8%A7%A3%E9%87%8A%E4%B8%80%E4%B8%8B%E4%B8%BA%E4%BB%80%E4%B9%88%EF%BC%9A%E2%88%AB0%E2%86%92%CF%80%2F2%EF%BC%9Asin%5E4%28x%29%3D%283%2F4%29%2A%281%2F2%29%2A%281%2F2%29%CF%80%E2%88%AB0%E2%86%92%CF%80%2F2%EF%BC%9Asin%5E6%28x%29%3D%285%2F6%29%2A%283%2F4%29%2A%281%2F2%29%2A%281%2F2%29%CF%80)
急!谁能求sin^4(x)的原函数?要过程.已知:F'(x)=sin^4(x)求:F(x)能否解释一下为什么:∫0→π/2:sin^4(x)=(3/4)*(1/2)*(1/2)π∫0→π/2:sin^6(x)=(5/6)*(3/4)*(1/2)*(1/2)π
急!谁能求sin^4(x)的原函数?要过程.
已知:F'(x)=sin^4(x)
求:F(x)
能否解释一下为什么:
∫0→π/2:sin^4(x)=(3/4)*(1/2)*(1/2)π
∫0→π/2:sin^6(x)=(5/6)*(3/4)*(1/2)*(1/2)π
急!谁能求sin^4(x)的原函数?要过程.已知:F'(x)=sin^4(x)求:F(x)能否解释一下为什么:∫0→π/2:sin^4(x)=(3/4)*(1/2)*(1/2)π∫0→π/2:sin^6(x)=(5/6)*(3/4)*(1/2)*(1/2)π
这个是要积分的
{积分号省略不写了}F'(x)dx
=(1-cos2x)/2*(1-cos2x)/2*dx
=(1-2cos2x+cos^2(2x))/4*dx
=(1-2cos2x+(1+cos4x)/2)/4*dx
=(3/8-cos2x/2+cos4x/8)dx
积分=(3/8)x-(sin2x)/2+(sin4x)/32
F(x)=1/COSx*(sinx)^5
F(x)=∫(sinx)^4dx
=∫[(sinx)^2(1-(cosx)^2)]dx
=∫[(sinx)^2-(sinx)^2(cosx)^2]dx
=∫[(sinx)^2-1/4*(sin2x)^2]dx
=∫[3/8-1/2*cos2x+1/8*cos4x]dx
=3/8x-1/4*sin2x+1/32*sin4x+C
一楼的解法有...
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F(x)=∫(sinx)^4dx
=∫[(sinx)^2(1-(cosx)^2)]dx
=∫[(sinx)^2-(sinx)^2(cosx)^2]dx
=∫[(sinx)^2-1/4*(sin2x)^2]dx
=∫[3/8-1/2*cos2x+1/8*cos4x]dx
=3/8x-1/4*sin2x+1/32*sin4x+C
一楼的解法有两处错误。
对于∫0→π/2:sin^4(x)
∫(sinx)^4dx知道了, 不就可以直接用牛顿莱布尼兹公式解了吗?结果是3π/16
对于0→π/2:sin^6(x),我有了一个解法,最好把你的邮箱告诉我,详细过程发给你。
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