令A=∫1/(tant+1)dt
=∫cost/(sint+cost)dt
B=∫sint/(sint+cost)dt
则A+B=∫cost/(sint+cost)dt+∫sint/(sint+cost)dt
=∫(cost+sint)/(sint+cost)dt
=∫dt=t+C1
A-B=∫cost/(sint+cost)dt+∫sint/(sint+cost)dt
=∫(cost-sint)/(sint+cost)dt
=∫d(sint+cost)/(sint+cost)
=ln|sint+cost|+C2
联立解得
A=1/2*[t+C1+ln|sint+cost|+C2]
=1/2*(t+ln|sint+cost|)+C
副产品
B=∫sint/(sint+cost)dt
=∫1/(cott+1)dt
=1/2*(t-ln|sint+cost|)+C
如果是∫[(1/tant)+1]dt
=∫(cott+1)dt
=∫cost/sintdt+t
=∫d(sint)/sint+t
=ln|sint|+t+C