RobertWolf
幼苗
共回答了17个问题采纳率:88.2% 举报
令 x=sint,则 dx=costdt,√(1-x²)=cost
设原积分为 A=∫[cost/(sint+cost)]dt
构造积分 B=∫[sint/(sint+cost)]dt
则
A+B=∫[(sint+cost)/(sint+cost)]dt=∫1dt=t+C1
A-B=∫[cost-sint/(sint+cost)]dt=∫d(sint+cost)/(sint+cost)=ln(sint+cost)+C2
解得
A=(1/2)[(A+B)+(A-B)]
=(1/2)[t+ln(sint+cost)]+C
其中 t=arcsinx
1年前
10