求不定积分∫[tan^2x/(1-sin^2x)]dx
问题描述：
求不定积分∫[tan^2x/(1-sin^2x)]dx
答
太简单,不必多说.
∫tan²x/(1-sin²x) dx
=∫tan²x/cos²x dx
=∫tan²x*sec²x dx
=∫tan²x d(tanx)
=(1/3)tan³x + C