1/((sinx)∧2+3(cosx)∧2)的不定积分
作者&投稿:舒全 (若有异议请与网页底部的电邮联系)
(cotx)'=(cosx/sinx)'=[(cosx)'*sinx-cosx*(sinx)']/(sinx)^2
=[-sinx*sinx-cosx*cosx]/(sinx)^2
= -1/(sinx)^2
所以
∫1/(sinx)^2 dx= -∫d(cotx)= -cotx +C
楼上确定把题目看对了吗?
迪古麻芩:[答案] 1,y=(cosx)^2-2acosx=(cosx-a)^2-a^2 0
隆尧县15218783523: (sinx/2)∧3的原函数 - ?
迪古麻芩: 答: ∫ sin³(x/2)dx =2∫ sin³(x/2) d(x/2) =- 2 ∫ [ 1-cos²(x/2)] d [cos(x/2)] =(2/3)cos³(x/2) -2cos(x/2)+C
隆尧县15218783523: 当x趋近于0时 求(1/xsin3x - xsin2/x)的极限 - ?
迪古麻芩: lim(x-->0)(tantanx-sinsinx)/x^3=lim(x-->0)(tanx-sinx)/x^3=lim(x-->0)(sec²x-cosx)/(3x^2)=lim(x-->0)[(1-2sec³x)sinx]/(6x)=1/6lim(x-->0)[6sec⁴xsin²x+(1-2sec³x)cosx]=1/6*(1-2...
隆尧县15218783523: 向量a=(2sinx,sinx)向量b=(cosx,2sinx)x属于实数,f(x)=向量a·向量b?
迪古麻芩: (1)f(x)=a*b=2sinxsinx+2√3sinxcosx=√3sin2x-cos2x+1=2(√3/2sin2x-1/2cos2x)=2cos(2x-π/6)+1考虑x属于[0,π/2],那么2x-π/6属于[-π/6,5π/6],那么最大值为3,当x=π/12即2x-π/6=0时取得,最小值为-√3/2,当x=π/2即2x-π/6=5π/6时取得
隆尧县15218783523: sinx+cosx/2sinx - cosx求tan=??
迪古麻芩: 设t=tanxt^2+2t-3=0(t+3)(t-1)=0t=-3或t=1∵x∈第二象限∴t=-3(2sinx-cosx)/(sinx+2cosx) [上下同时除于cosx]=(2tanx-1)/(tanx+2)=[2*(-3)-1]/(-3+2)=(-7)/(-1)=7
隆尧县15218783523: ∫1 - 1 (5x∧3 - 3x+2)dx= - ?
迪古麻芩: ∫1 -1 (5x∧3-3x+2)dx=(5/4)x∧4-(3/2)x∧2+2x |1 -1=(5/4)[1^4-(-1)^4]-(3/2)[1^2-(-1)^2]+2[1-(-1)]=4
隆尧县15218783523: ∫(tanx∧2+cosx∧3)dx - ?
迪古麻芩: 解:∫((tanx)^2+(cosx)^3)dx=∫(1/(cosx)^2-1)dx+∫(1-(sinx)^2)d(sinx)=tanx-x+sinx-(1/3)(sinx)^3+C 所以 ∫((tanx)^2+(cosx)^3)dx=(-1/3)(sinx)^3+tanx+sinx-x+C 希望能帮到你!
隆尧县15218783523: 若tanx=√2 求2(sinx)∧2 - sinxcosx+(cosx)∧2的值 - ?
迪古麻芩:[答案] 2(sinx)∧2-sinxcosx+(cosx)∧2 =(2(sinx)∧2-sinxcosx+(cosx)∧2)/[sinx)∧2+(cosx)∧2] =(2(tanx)^2-tanx+1]/((tanx)^2+1) =(5-√2)/3
隆尧县15218783523: 如果(x3)/2 - (x - 8)/6与1 - (21)/3互为相反数,求方程a - 3=ax中a的值?
迪古麻芩: 解:∵x+3/2-x-8/6与1-2x+1/3互为相反数, ∴ x+3/2-x-8/6﹢1-2x+1/3=0 ∴-2X=-3/2 ∴X=3/4∴ax-3=a+x﹙3/4﹚a-3=a+3/4 ﹙-1/4﹚a=15/4 a=-15
隆尧县15218783523: 求∫1/x^3+1的不定积分 - ?
迪古麻芩: 解答过程如下: 先分解因式: ∫ 1/(x³ + 1) dx = ∫ 1/[(x + 1)(x² - x + 1)] dx = ∫ A/(x + 1) dx + ∫ (Bx + C)/(x² - x + 1) dx 1 = A(x² - x + 1) + (Bx + C)(x + 1) = Ax² - Ax + A + Bx² + Cx + Bx + C 1 = (A + B)x² + (- A + B + C)x + (A + C) { A + B = 0 { - A + ...