Question

数学积分题求解 求(3-sinx)/(3+cosx)的不定积分

Answer 1

原式
=∫3/(3+cosx)dx - ∫sinx/(3+cosx)dx
=∫{3[sin(x/2)]^2+3[cos(x/2)]^2} /4[sin(x/2)]^2+4[cos(x/2)]^2-2[sin(2/x)]^2 dx + ∫1/(3+cosx)d(3+cosx)
=∫{3[tan(x/2)]^2+3 / 2[tan(x/2)]^2+4}dx + ln(3+cosx) + C
设tan(x/2)=t,则x=2arctan(t)
=∫[3(n^2+1)/(2n^2+4) * 2/(n^2+1)]dx + ln(3+cosx) + C
=∫3/(2+n^2)dx + ln(3+cosx) + C
=(3/2)∫1/1+(n/√2)^2 + ln(3+cosx) + C
=(3/2)arctan[tan(x/2)/√2] + ln(3+cosx) + C
PS：前边那个回答不可信.