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证明定积分(0到π/2)sin^3x/(sinx+Cosx)Dx=定...

解:∵(cosx)^4是偶函数,(sinx)^3是奇函数 ∴∫(cosx)^4dx=2∫(cosx)^4dx ∫(sinx)^3dx=0 故 ∫((cosx)^4+(sinx)^3)dx =∫(cosx)^4dx+∫(sinx)^3dx =2∫(cosx)^4 =(1/2)∫[3/2+2cos(2x)+cos(4x)/2]dx (应用倍角公式) =(1/2)[3x/2+sin(2x)+sin(4x)/8]│ =(1...

∫π/2 0 (cos2x/cosx+sinx)dx =∫π/2 0 (cos²x-sin²x)/(cosx+sinx)dx =∫π/2 0 (cosx-sinx)dx =sinx+cosx π/2 0 =(1+0)-(0+1) =0

f'(x)=3sin^2 x cosx-3cos^2 x sinx=3sinxcosx(sinx-cosx)=1.5sin2x(sinx-cosx) 得sinx=0或cosx=0或 sinx-cosx=0 得极值点x=π, π/2, 3π/2, π/4, 5π/4 检验极值点左右邻域的符号,左减右加才为极小值,符合的有: x=π,3π/2, π/4 其有3个极小值点

1. sinX-cosX=1/5 1-2sinxcosx=1/25 sinxcosx=12/25 2. (sinx+cosx)²=1+2sinxcosx=1+24/25=49/25 因为sinX-cosX=1/5,0

积分:1/sinxdx =积分:1/(2sinx/2cosx/2)dx =1/2积分:(sinx/2^2+cosx/2^2)/(sinx/2cosx/2)dx =1/2积分:(tanx/2+cotx/2)dx =1/2*[(-2)ln|cosx/2|+2ln|sinx/2|)+C =ln|sinx/2|-ln|cosx/2|+C =ln|tanx/2|+C

原式=sinx-(1/3)sin3x ;区间为-π/2 到π/2 =sin(π/2)-(1/3)sin3π/2-[sin(-π/2)-(1/3)sin(-3π/2)] =1+1/3-(-1-1/3) =2+2/3 =2又2/3

(x+sin(x))/(4cos(x)-4) 首先改写降次: xcos^4(x/2)/sin^3(x) = 1/8 xcot(x/2)csc^2(x/2) ... 注:csc(x)=1/sin(x) 换元积分:令u=x/2: dx=2du, ∫ 1/8 xcot(u)csc^2(u) 2du =1/8 * 2 * 2 ∫ ucot(u)csc^2(u) du 然后分部积分, =1/2 u(-1/2cot^2...

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