Question

不定积分题，求解，谢谢。

Answer 1

∫[1/(a²sin²x+b²cos²x)]dx
=∫[(1/cos²x)/(a²sin²x/cos²x)+b²]dx
=∫[sec²x/(a²tan²x+b²)]dx
=∫[1/(a²tan²x+b²)]d(tanx)
=(1/ab)∫[1/(a/b)tanx)+1]d((a/b)tanx)
=(1/ab)arctan((a/b)tanx)+C