Question

关于微积分的几个题

Answer 1

1. ∫[cos2x/cos²xsin²x]dx=∫[(2cos²x-1)/cos²xsin²x]dx=2∫(1/sin²x)dx-(1/4)∫(1/sin²2x)dx

   =-2cotx+(1/8)cot2x+c

2. ∫xcosxdx=∫xd(sinx)=xsinx-∫sinxdx=xsinx+cosx+c

3. ∫【-2,2】[x²√(4-x²)]dx=2∫【-2,2】{x²√[1-(x/2)²]}dx

   【令x/2=sinu,则x=2sinu，dx=2cosudu；x=-2时u=-π/2；x=2时u=π/2】

   =8∫【-π/2.π/2】sin²ucosudu=8∫【-π/2，π/2】sin²ud(sinu)=[(8/3)sin³u]【-π/2，π/2】

   =(8/3)(1+1)=16/3;

4. ∬【D】e^(-x²-y²)dxdy=∫e^(-x²)dx∫e^(-y²)dy【请给出积分域D，不然无法求解】

5. ∬【D】[(siny)/y]dxdy=∫【0,1】dx∫【x，1】[(siny)/y]dy【令siny=u，则y=arcsinu；

   dy=du/√(1-u²)；y=x时u=sinx；y=1时u=sin1】

   =∫【0,1】dx∫【sinx，sin1】[u/√(1-u²)]du

   =∫【0,1】dx[-(1/2)∫【sinx，sin1】d(1-u²)/√(1-u²)]

   =-(1/2)∫【0,1】[2√(1-u²)]【sinx，sin1】dx

   =-[∫【0,1】[√（1-sin²1)-√(1-sin²x)]dx=-[∫【0,1】[cos1-cosx]dx

   =-[xcos1-sinx]【0,1】=[sinx-xcos1]【0,1】=sin1-cos1.