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(Cosx)^4求0到π/4的定积分

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)/...

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: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。

cos^4(x/2) =[cos^2(x/2)]^2 =[(1+cosx)/2]^2 =(1/4)(1+2cosx+cos^2x) =(1/4)[1+2cosx+(1+cos2x)/2] =(1/4)(3/2+2cosx+1/2*cos2x) 积分得 x-(1/4)(3x/2+2sinx+1/4*sin2x) 代入上下限得 π/4-(1/4)(3/2*π/4+2*√2/2+1/4*1) =(1/32)(5π-2-8√2)

点击放大:

公式:∫[0→π] xf(sinx) dx = (π/2)∫[0→π] f(sinx) dx ∫[0→π] x(sinx)⁶(cosx)⁴ dx 由公式: =(π/2)∫[0→π] (sinx)⁶(cosx)⁴ dx =(π/2)∫[0→π/2] (sinx)⁶(cosx)⁴ dx + (π/2)∫[π/2→π] (sinx)⁶(cosx)⁴...

同样的方法可求(cosx)^4的积分。

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

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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