第1个回答 2014-01-06
(1)分部积分法:
∫arcsinxdx =x*arcsinx -∫xdarcsinx =x*arcsinx -根号(1-x^2) +c
(2)分部积分法:
∫e^x sin^2 x dx =∫sin^2 x de^x =e^x*sin^2 x -∫e^x dsin^2 x
=e^x *sin^2 x -∫2sinxcosx e^x dx
=e^x *sin^2 x -∫sin2x e^x dx ...(i)
=e^x *sin^2 x -∫sin2x de^x
=e^x *sin^2 x -e^x *sin2x +∫e^x dsin2x
=e^x*sin^2 x -e^x *sin2x +∫e^x*cos2x*2dx
=e^x*sin^2x -e^x*sin2x+∫2cos2x de^x
=e^x*sin^2 x -e^x *sin2x +2cos2x*e^x -∫2e^xdcos2x
=e^x*sin^2 x-e^x*sin2x+2cos2x*e^x +∫4sin2x e^xdx ....(ii)
注意到(i) (ii)行可以求得∫sin2x e^x dx =1/5(e^x *sin2x -2cos2x *e^x )
所以∫e^x *sin^2 x dx=e^x *sin^2 x -1/5(e^x *sin2x -2cos2x*e^x) +c
(3)换元法:
∫1/(x^2+2x+2)dx =∫1/((x+1)^2+1) dx (令x+1=tana)
=∫1/tan^2 a+1) dtana
=ln lx/(x+2)l /2 /(-无穷大,+无穷大)=0
(4)凑微分法
∫1/(e^(2+x)+e^(3-x) )dx
=∫1/(e^2*e^x+e^3/e^x)dx
=∫1/e^2 *e^x/(e^(2x)+e) dx
=1/e^2 ∫1/(e^(2x)+e) de^x (令e^x =t)
=1/e^2 ∫1/(t^2+e)dt
=1/e^2 *ln l( e^x -e^(1/2)) /(e^x+e^(1/2) l /(-无穷大,+无穷大)
=0本回答被网友采纳