求不定积分~

2025-04-23 17:36:22
推荐回答(1个)
回答1:

令x=sint,则dx=costdt
原式=∫cost/(sint+cost)dt
令A=∫cost/(sint+cost)dt,B=∫sint/(sint+cost)dt
则A+B=∫(sint+cost)/(sint+cost)dt=∫dt=t+C
A-B=∫(cost-sint)/(sint+cost)dt=∫d(sint+cost)/(sint+cost)=ln|sint+cost|+C
两式联立,得:A=[t+ln|sint+cost|]/2+C
原式=[arcsinx+ln|x+√(1-x^2)|]/2+C,其中C是任意常数