1cost3的定积分

敖沫瑞849求定积分2π∫ - 0^2π - (√2)a^2(1 - cost)^(3/2) dt
胥章查13188158523 ______ 2π∫(0→2π) √2a²(1 - cost)^(3/2) dt = 2√2πa²∫(0→2π) √2a²[2sin²(t/2)]^(3/2) dt,cosx = 1 - 2sin²(x/2) => 1 - cosx = 2sin²(x/2) = 2√2πa²∫(0→2π) 2^(3/2) * |sin³(t/2)| dt = 16πa²∫(0→2π) sin³(t/2) d(t/2),在t∈[0,2π],sin³(t/2) ≥ 0,∴|sin³(t/...

敖沫瑞849紧急求助,解积分,∫ln(1+x^2)dx,要过程,在线等 -
胥章查13188158523 ______ 分部积分 ∫ln(1+x^2)dx,= x ln(1+x^2) - ∫[x 1/(1+x^2) 2x] dx= x ln(1+x^2) - 2 [∫[1 - 1/(1+x^2)] dx]= x ln(1+x^2) - 2 x + 2 arctan(x) + C

敖沫瑞849(cost+sint)^4的积分 -
胥章查13188158523 ______[答案] ∫[(cost+sint)^4]dt = ∫{(cost)^4+4[(cost)^3]sint+6(costsint)^2+4cost[(sint)^3]+(sint)^4}dt = …… (按三角函数积分的方法计算,留给你)

敖沫瑞849定积分:分母是X乘上根号下(x^2+4),分子是一.上限是2根号3,下限是2 -
胥章查13188158523 ______ 换元法:x=2tanθ,由于x∈[2,2√3],从而θ∈[π/4,π/3],√(x²+4)=2/cosθ, dx=2dθ/cos²θ,从而∫[2,2√3]dx/(x√(x²+4))=1/2∫[π/4,π/3]dθ/sinθ=-1/4*ln2+1/2*ln(2+√2)-1/4*ln3 凑微分法:∫[2,2√3]dx/(x√(x²+4))=∫[2,2√3]dx/(x²√(1+4/x²))=∫[2,2√3]x^(-2)dx/√(1+4/x²)=-1/2∫[2,2√3]d(2/x)/√(1+(2/x)²)=-1/2ln(2/x+√(1+(2/x)²))[2,2√3]=-1/4*ln2+1/2*ln(2+√2)-1/4*ln3

敖沫瑞849求定积分2π∫ - 0^2π - (√2)a^2(1 - cost)^(3/2) dt要过程啊,最好说明用了什么公式答案是(64/3)πa^2 -
胥章查13188158523 ______[答案] 2π∫(0→2π) √2a²(1 - cost)^(3/2) dt= 2√2πa²∫(0→2π) √2a²[2sin²(t/2)]^(3/2) dt,cosx = 1 - 2sin²(x/2) => 1 - cosx = 2sin²(x/2)= 2√2πa²∫(0→2π) 2^(3/2) * |s...

敖沫瑞849cosxsinx^3的定积分b -
胥章查13188158523 ______ ∫cosxsinx^3dx=∫sin³xdsinx=1/4sin^4x+c

敖沫瑞849求t/(1+cost)的不定积分求定积分t/(1+cost),上限π/2,下限是2π/3 -
胥章查13188158523 ______[答案] ∫t/(1+cost) dt =2∫t [sec(t/2)]^2 dt =∫ tdtan(t/2) = ttan(t/2) - ∫ tan(t/2) dt = ttan(t/2) - ∫ sin(t/2)/cos(t/2)dt = ttan(t/2) + 2 ln|cos(t/2)|+ C

敖沫瑞849(1 - cost)2转化为sint -
胥章查13188158523 ______[答案] 解(1-cost)2 =(1-(1-2sin^2(t/2)))2 =(2sin^2(t/2))2 =4sin^4(t/2).

敖沫瑞849请问sint的4次方乘以cost的平方的积分怎么求? -
胥章查13188158523 ______[答案] =(1/4)(1/2)(1/2) ∫(1-cos4t)(1-cos2t)dt =(1/16) ∫(1-cos4t-cos2t+cos4tcos2t)dt =t/16-(sin4t)/64-(sin2t)/32+(1/16)(1/2)∫(cos6t+cos2t)dt =t/16-(sin4t)/64-(sin2t)/32+(1/192)*(sin6t)+(sin2t)/64+C