1cost3化简

盖曼娅1752求定积分∫(1 - √3)dx/(x√(x^2+1)) -
冷显郑15011022155 ______ 令x=tana dx=sec²ada x=√3,a=π/3 x=1,a=π/4 原式=∫(π/4,π/3)sec²ada/(tanaseca)=∫(π/4,π/3)da/sina=∫(π/4,π/3)sinada/(1-cos²a)=-∫(π/4,π/3)dcosa/(1-cosa)(1+cosa)=-1/2∫(π/4,π/3)[-1/(cosa-1)+1/(1+cosa)]dcosa=-1/2ln[(1+cosa)/(1-cosa)](π/4,π/3)=-1/2ln(1+1/2)/(1-1/2)+1/2ln(1+√2/2)/(1-√2/2)=ln[(√6+√3)/3]

盖曼娅1752定积分(+∞到1)∫1/x(x^2+1)dx -
冷显郑15011022155 ______ =(+∞到1)∫(x+1/x)dx=[(x^2)/2 +lnx] | (+∞到1)=+∞

盖曼娅1752(3分之1)的负1次方怎么化简
冷显郑15011022155 ______ (3分之1)的1次方的倒数,即是3

盖曼娅1752sintsin(2t)怎样化简为1/2cost+1/2cos(3t)步骤,谢谢 -
冷显郑15011022155 ______[答案] 答:结论不对,后面应该是减号 cost-cos(3t) =cos(2t-t)-cos(2t+t) =cos(2t)cost+sin(2t)sint- [ cos(2t)cost-sin(2t)sint ] =2sintsin(2t) 所以: sintsin(2t)=(1/2)cost -(1/2)cos(3t)

盖曼娅1752如图,在直角梯形ABCD中,AB//DC,DA垂直AB,且AD=DC=1,AB=3若点P在以C为圆心,与直线B -
冷显郑15011022155 ______ 以点A为原点,AB和AD分别为x,y轴建立直角坐标系 则B,C,D坐标分别为(3,0),(1,1),(0,3) |BD|=√(1+9)=√10,设C到BD的高为h 则有△DCB面积求法可知h|BD|=|CD|·|AD| ∴h=1/√10,∴P在圆(x-1)²+(y-1)²=1/10 又向量AP=a向量AD...

盖曼娅1752点M(4,3)点P(x,y)为圆M上任意一点,以M为圆心,半径为1做圆求y - x最大值 (提示用y=x) -
冷显郑15011022155 ______ 圆M:(x-4)^+(y-3)^=1,点P(x,y)为圆M上任意一点,∴x=4+cosa,y=3+sina,∴y-x=-1+sina-cosa =-1+√2sin(x-π/4),当a=3π/4时y-x取最大值√2-1.

盖曼娅1752一又三分之一化简是多好 -
冷显郑15011022155 ______ 一又三分之一化简是多好 1又1/3就是最简带分数形式 化成假分数是4/3

盖曼娅17521比4分之3化简 -
冷显郑15011022155 ______ 1比4分之3=4/3

盖曼娅1752t属于(0,π),sint+cost=1/3,求cos2t -
冷显郑15011022155 ______[答案] ∵(sint+cost)^2 =1+2sintcost =1/9 ∴sintcost=-4/9 ∵t∈(0,π) ∴sint>0 ∵sintcost

盖曼娅1752sint/(1 - cost)求导 -
冷显郑15011022155 ______[答案] f'(t)=[(sint)'(1-cost)-(sint)(1-cost)']/(1-cost)? =[cost(1-cost)-sintsint]/(1-cost)? =[cost-(cos?t+sin?t)]/(1-cost)? =(cost-1)/(1-cost)? =1/(cost-1)