原積分可化為∫√(25-(x+3)^2)dx
用第二換元法,設(shè)x+3=5sint,t∈[-π/2,π/2],則d(x+3)=5costdt
原積分化為
∫√(25-(x+3)^2)dx
=∫√(25-(x+3)^2)d（x+3）
=∫√(25-25(sint)^2)*5costdt
=∫5*|cost|5costdt
=25∫(cost)^2dt t∈[-π/2,π/2]=>cost>0
=25/2∫(cos2t-1)dt
=25/2(sin2t/2-t)+C(C為任意常數(shù))
=25sin2t/4-25t/2+C(C為任意常數(shù))
t=arcsin((x+3)/5)
sint=(x+3)/5,cost=√(1-（(x+3)/5）……2)=√（16-6x-x^2）/5
得,原積分=（x+3）√（16-6x-x^2）/2+25arcsin((x+3)/5)/2+C