求不定积分∫1/(a^2-sinx^2)dx

1[a^2-(sinx)^2]=2(2a^2-1+cos2x)
t=tanx dx=dt(t^2+1) 2dx(2a^2-1+cos2x)
=2dt[(2a^2-1)(t^2+1)+1-t^2]=dt[(a^2-1)t^2+a^2]
|a|>1 原式=∫(1a^2)dt[(a^2-1)t^2a^2+1]
=1[|a|(a^2-1)^(12)]*arctan[(a^2-1)^(12)t|a|]+C
=1[|a|(a^2-1)^(12)]*arctan[(a^2-1)^(12)tanx|a|]+C
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