Question

求不定积分：积分符号(sinx/cosx三次方)dx，我尝试用2种方法结果有两个答案，过程都觉得对

Answer 1

设tan(x/2)=t
则sin
x=2t/(1+t^2)
cos
x=(1-t^2)/(1+t^2)
dx=2/(1+t^2)dt
∫sinx/(sinx+cosx)dx=∫2t/(1+t^2)*2/(1+t^2)/[2t/(1+t^2)
+(1-t^2)/(1+t^2)]dt
=-4∫t/[(1+t^2)(t+√2-1)(t-√2-1)dt
设：
t/[(1+t^2)(t+√2-1)(t-√2-1)=a/(t-√2-1)+b/(t+√2-1)+(ct+d)/(1+t^2)
a(t+√2-1)(1+t^2)+b(t-√2-1)(1+t^2)+(ct+d)(t+√2-1)(t-√2-1)=t
t=1-√2:
-2√2(4-2√2)b=1-√2
b=1/8
t=1+√2:
2√2(4+2√2)a=1+√2
a=1/8
t=0:
1/8(√2-1)+1/8(-√2-1)+d(√2-1)(-√2-1)=0
d=-1/4
t=1:
1/8(1+√2-1)(1+1)+1/8(1-√2-1)(1+1)+(c-1/4)(1+√2-1)(1-√2-1)=1
c=-1/4
原式=-4∫[1/8/(t-√2-1)+1/8/(t+√2-1)-1/4(t+1)/(1+t^2)]dt
=-1/2∫dt/(t-√2-1)-1/2∫dt/(t+√2-1)+∫t/(1+t^2)dt+∫dt/(1+t^2)
=-1/2ln|t-√2-1|-1/2ln|t+√2-1|+1/2ln|1+t^2|+arctan
t
+c
=-1/2ln[(t-√2-1)(t+√2-1)/(1+t^2)]+arctan
t
+c
=
x/2-1/2*(ln(sinx+cosx))+c
工作量太大了，采纳并加分！！