这个用什么方法呢？ 第1页

defd{mathrm d} egin{align} int^0_{-1}(x+1)sqrt{1-x-x^2}d x &=int^0_{-1}(x+1)sqrt{frac54-left(x+frac12 ight)^2}d x\ &(t=x+frac12,d t=d x)\ &=int^frac12_{-frac12}(t+frac12)sqrt{frac54-t^2}d t\ &(frac{sqrt5}2sin u=t,frac{sqrt5}2cos ud u=d t,u=arcsinleft(frac2{sqrt5}t ight))\ &=int^{arcsinfrac1{sqrt5}}_{-arcsinfrac1{sqrt5}}(frac{sqrt5}2sin u+frac12)sqrt{frac54-frac54sin^2u}frac{sqrt5}2cos ud u\ &=frac58int^{arcsinfrac1{sqrt5}}_{-arcsinfrac1{sqrt5}}(sqrt5sin u+1)cos^2ud u\ hline int(asin u+b)cos^2ud u &=int(asin u+b)(1-sin^2u)d u\ &=int(asin u+b-asin^3u-bsin^2u)d u\ &=-acos u+bu-aintsin^3ud u-bintsin^2ud u\ &=-acos u+bu+aint(1-cos^2u)dcos u-bintfrac{1-cos2u}2d u\ &=-acos u+bu+aleft(cos u-frac13cos^3u ight)+bleft(frac14sin2u-frac12u ight)+C\ &=frac b2u-frac a3cos^3u+frac b4sin2u+C\ hline &=frac58left(frac u2-frac{sqrt5}3cos^3u+frac14sin2u ight)^{arcsinfrac1{sqrt5}}_{-arcsinfrac1{sqrt5}}\ &=frac14+frac58arcsinfrac1{sqrt5} end{align}\