Question

求不定积分∫[sinxcosx/(sinx+cosx)]dx

Answer 1

∫[sinxcosx/(sinx+cosx)]dx=-1/4∫[dcos2x/(sinx+cosx)]
=-1/4cos2x/(sinx+cosx)-1/4/∫[cos2x*(cosx-sinx)/(sinx+cosx)^2dx
=-1/4cos2x/(sinx+cosx)-1/4∫[(cosx-sinx)^2/(sinx+cosx)]dx
=-1/4cos2x/(sinx+cosx)-1/4∫[(1-2sinx*cosx)/(sinx+cosx)]dx
=-1/4cos2x/(sinx+cosx)-1/4∫[1/(sinx+cosx)]dx+1/2∫[sinxcosx/(sinx+cosx)]dx
所以
∫[sinxcosx/(sinx+cosx)]dx=2*【-1/4cos2x/(sinx+cosx)-1/4∫[1/(sinx+cosx)]dx】
=-1/2cos2x/(sinx+cosx)-1/2∫[1/(sinx+cosx)]dx

∫dx/(sinx+cosx),令u=tan(x/2),dx=2du/(1+u²)
=∫1/[2u/(1+u²)+(1-u²)/(1+u²)]·2/(1+u²)] du
=2∫1/[(2u+1-u²)/(1+u²)·1/(1+u²)] du
=2∫1/(2u+1-u²) du
=-2∫1/(u²-2u-1) du
=-2∫1/[u²-2u+(2/2)²-1-(2/2)²] du
=2∫1/[2-(u-1)²] du,令A=u-1,dA=du
=2∫1/(2-A²) du,用三角代换,令A=√2sinT,dA=√2cosTdT
=2∫[1/(2-2sin²T)·√2cosT]dT
=2√2∫cosT/(2cos²T) dT
=√2∫secT dT
=√2ln|secT+tanT|+C
sinT=A/√2,r=√2,y=A.x=√(2-A²)
secT=r/x=√2/√(2-A²)
=√2ln|(√2+A)/√(2-A²)|+C
=√2ln|(√2+u-1)/√[2-(u-1)²]|+C
=√2ln|[√2+tan(x/2)-1]/√{2-[tan(x/2)-1]²}]|+C
所以∫[sinxcosx/(sinx+cosx)]dx
=-1/2cos2x/(sinx+cosx)-1/2∫[1/(sinx+cosx)]dx
=-1/2cos2x/(sinx+cosx)-√2/2ln|[√2+tan(x/2)-1]/√{2-[tan(x/2)-1]²}]|+C

太太太麻烦了

Answer 2

换元，令u＝tan(x/2)，则由三角函数的万能公式，sinx 和 cosx 都能写成u的形式 ，那么原积分可以化成有理函数的积分，然后再考虑真分式拆分之类的

Answer 3

∫[sinxcosx/(sinx+cosx)]dx
=∫1/2sin(2x)(cosx-sinx)/cos(2x)dx
=2^0.5/4∫tan(2x)cos(x+Pi/4)dx
=2^0.5/4∫tan(2x)dsin(x+Pi/4)
=2^0.5/4 tan(2x)sin(x+Pi/4)-∫sin(x+Pi/4)dtan(2x)
=...

Answer 4

=(tan(x/2) - 1)/(tan(x/2)^2 + 1) + (2^(1/2)*arctan((12*2^(1/2)*tan(x/2)*i + 4*2^(1/2)*i)/(16*tan(x/2) + 8))*i)/2+C
这是matlab命令求出的结果，我也不知道中间的过程