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定积分∫(0到π/4)(Cosx)^4=

解: ∫【0→π/4】(cosx)^4dx =∫【0→π/4】[(cos2x+1)/2]²dx =∫【0→π/4】(cos²2x+2cos2x+1)/4 dx =1/4 ∫【0→π/4】[(cos4x+1)/2+2cos2x+1]dx =1/4 ∫【0→π/4】[(cos4x)/2+2cos2x+3/2]dx =【0→π/4】1/4 [(sin4x)/8+sin2x+3x/2] =1/4[(sinπ)/...

解:分享一种解法。∵tanxdx=sinxdx/cosx=-d(cosx)/cosx=-d[ln(cosx)], ∴原式=-∫(0,π/4)ln(cosx)d[ln(cosx]=-(1/2)[ln(cosx)]^2丨(x=0,π/4)=-(1/8)(ln2)^2。供参考。

点击放大:

华里士公式,是公式,可以记住,也可以去搜搜看看怎么推的

解: ∫cos⁴xdx =∫(cos²x)²dx =∫[(1+cos2x)/2]²dx =(1/4)∫(1+cos²2x+2cos2x)dx =(1/4)∫[1+(1+cos4x)/2+2cos2x]dx =(1/4)[x+x/2+(sin4x)/8+sin2x]+C

具体步骤如下: (cosx)^4=cos⁴x=(cos²x)²=[(1+cos2x)/2]²=(1/4)(1+2cos2x+cos²2x)=(1/4)+(1/2)cos2x+(1/8)(1+cos4x)=(3/8)+(1/2)cos2x+(1/8)cos4x∫cos⁴xdx=∫[(3/8)+(1/2)cos2x+(1/8)cos4x]dx=(3/8)x+(1/4)sin2x...

解:f(x)=asinx+bcosx∫【x=0→π/2】f(x)dx=∫【x=0→π/2】(asinx+bcosx)dx=a∫【x=0→π/2】sinxdx+b∫【x=0→π/2】cosxdx=a【x=0→π/2】-cosx+b【x=0→π/2】sinxdx=a[-cos(π/2)+cos0]+b(sin(π/2)-sin0)=a+b依题意,有:a+b=4……………………………………………………(1)∫【x=...

∫ cosx^4dx=∫ cosx^2dsinx^2=∫ (1-sinx^2)dsinx^2 =∫ dsinx^2-∫sinx^2 dsinx^2=∫cosx^2 dx-(sinx^3)/3 =1/4∫(cos2x+1) d2x-(sinx^3)/3 =sin2x/4+x/2-(sinx^3)/3

解:∵(sinx)^4(cosx)^4=[sin(2x)/2]^4 =[sin²(2x)]²/2^4 =[(1-cos(4x))/2]²/2^4 =[1-2cos(4x)+cos²(4x)]/2^6 =[1-2cos(4x)+(1+cos(8x))/2]/2^6 =[3-4cos(4x)+cos(8x)]/2^7 ∴∫(sinx)^4(cosx)^4dx=(1/2^7)∫[3-4cos(4x)+cos(8x...

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