Question

求教高数不定积分…

求Sin x倒数的不定积分...................希望能详细说明,谢谢.

Answer 1

∫ 1 / sin x dx = 1/2 ∫ 1 / (sin[x/2] * cos[x/2]) dx = 1/2 * ∫( (sin[x])^2 + (cos[x])^2 ) / (sin[x/2] * cos[x/2])dx
= 1/2 * [ ∫ sin[x/2] / cos[x/2] dx + ∫ cos[x/2] / sin[x/2] dx ]
= ∫ sin[x/2] / cos[x/2] d(x/2) + ∫ cos[x/2] / sin[x/2] d(x/2)
= - ∫ 1 / cos[x/2] d(cos[x/2]) + ∫ 1 / sin[x/2] d(sin[x/2])
= - ln | cos[x/2] | + ln | sin[x/2] | + C
= ln| tan[x/2] | + C
C为任意常数.