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.需要详细思路,最好拍照上传.

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.需要详细思路,最好拍照上传.

1=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+c2∫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)+c3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）=π4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}=4/56设t=1+√3x+1 ,2

谢谢了