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題目應該是說求(cosx)^2/(sinx)^2的不定積分.設t=tanx,則x=arctant,dx=dt/(1+t²)∫(cosx)^2/(sinx)^2dx=∫dt/[t²(t²+1)] =∫[1/t²-1/(t²+1)]dt=-1/t-arctant+C,再將t=tanx帶回來,得...
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