利用凑微分法，换元法，分部积分法计算不定积分，定积分和广义积分。 广义积分能用分部积分法求解吗？
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\u222bx/\u221a1-x² dx =\u222b1/2\u221a1-x² dx ²
=-\u222b1/2\u221a1-x² d(1-x²)
=-\u221a1-x² \u518d\u628a0\u30011\u5e26\u5165\u5c31\u884c\u4e86
=xarcsinx-∫x/[(1-x^2)^1/2]dx=xarcsinx+1/2*∫d(1-x^2)/[(1-x^2)^1/2]=xarcsinx+(1-x^2)^1/2+c
2
∫e^xsin^2xdx=∫(1-cos2x)e^x/2dx=1/2[∫e^xdx-∫e^xcos2xdx]
下面着重求出第二项
∫e^xcos2xdx=∫cos2xd(e^x)=e^xcos2x+2∫e^xsin2xdx=e^xcos2x+2∫sin2xde^x
=e^xcos2x+2e^xsin2x-4∫e^xcos2xdx
移项得到
5∫e^xcos2xdx=e^xcos2x+2e^xsin2x
所以∫e^xcos2xdx=1/5(e^xcos2x+2e^xsin2x)
代入原式得到
∫e^xsin^2xdx=1/2[e^x-1/5(e^xcos2x+2e^xsin2x)]=e^x(1/2-1/10cos2x-1/5sin2x)+c
3
原式=∫{-无穷到+无穷}d(x+1)/[1+(x+1)^2]=arctan(x+1)|{-无穷到+无穷}=π/2-(-π/2）=π
4
原式=∫e^(-5/2)d[e^(x-1/2)]/[1+[e^(x-1/2)]^2]=e^(-5/2)arctan[e^(x-1/2)] |{负无穷到正无穷}=π/2*(e^(-5/2))
5
原式=∫√sin^(3)x (1-sin^(2)x) dx=∫sin^(3/2)x |cosx|dx
=∫{0到π/2}sin^(3/2)x cosxdx-∫{π/2到π}sin^(3/2)x cosxdx
=∫{0到π/2}sin^(3/2)xdsinx-∫{π/2到π}sin^(3/2)xdsinx
=2/5(sin^(5/2)x)| {0到π/2}-2/5(sin^(5/2)x)| {π/2到π}
=4/5
6
设t=1+√3x+1 ，2<t<5
那么x=1/3 [(t-1)^2-1]
所以dx=2/3 (t-1) dt
那么
原式=2/3 ∫{t从2到5}[(t-1)/t]dt
=2/3 ∫{t从2到5}[(1-1/t)]dt
=2/3(t-lnt) | {t从2到5}
=2-2/3 ln(5/2)
(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
绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細
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