Question

1+cosx/1+sinx的平方的不定积分

Answer 1

原式=∫(1/2-cos^2)dx+∫(1/(1+sin^2x))d(sinx)=∫[1/(cos^2(2sec^2x-1))]dx+arctan(sinx)=∫(1/(1+2tan^2x))d(tanx)+arctan(sinx)=[(√2)/2]arctan[(√2)(tanx)]+arctan(sinx)+C