∫dt/sint等于多少,应该看成对数函数的求导还是幂函数的求导

Question

∫dt/sint 等于多少,应该看成对数函数的求导还是幂函数的求导

Answer 1

∫dt/sint
= ∫sintdt/sin²t
= -∫dcost/sin²t
= -∫dcost/(1-cos²t)
= -∫dcost/(1-cost)(1+cost)
= -½∫[1/(1-cost) + 1/(1+cost)]dcost
= -½∫dcost/(1-cost) -½∫dcost/(1+cost)
= ½∫d(1-cost)/(1-cost) -½∫d(1+cost)/(1+cost)
= ½ln|1-cost| - ½ln|1+cost| + C
= ½ln|(1-cost)/(1+cost)| + C
= ½ln|2sin²(t/2)/2cos²(t/2)| + C
= ½ln|tan²(t/2)| + C
= ln|tan(t/2)| + C