∫1/(sinx+cosx)dx ∫sinx/(1+sinx)dx ∫1/(3+cosx)dx ∫ 1/(1+sinx+cosx)dx
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∫1/(sinx+cosx)dx ∫sinx/(1+sinx)dx ∫1/(3+cosx)dx ∫ 1/(1+sinx+cosx)dx
∫1/(sinx+cosx)dx ∫sinx/(1+sinx)dx ∫1/(3+cosx)dx ∫ 1/(1+sinx+cosx)dx
∫1/(sinx+cosx)dx ∫sinx/(1+sinx)dx ∫1/(3+cosx)dx ∫ 1/(1+sinx+cosx)dx
基本上4条都用万能公式代换
首先令u=tan(x/2),那么du=(1/2)sec²(x/2)dx
du=2du/(1+u²),sinx=2u/(1+u²),cosx=(1-u²)/(1+u²)
第一条:
∫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
第二条:
∫sinx/(1+sinx) dx,令u=tan(x/2),dx=2du/(1+u²)
=∫[1/[1+2u/(1+u²)]·2u/(1+u²)·2/(1+u²)] du
=4∫u/(u^4+2u³+2u²+2u+1) du
=4∫u/[(u+1)²(u²+1)] du
=4∫{1/[2(1+u²)-1/[2(1+u)²]} du
=2∫du/(1+u²)-2∫du/(1+u)²
=2arctan(u)-2∫d(1+u)/(1+u)²
=2arctan(u)+2/(1+u)+C
=2arctan[tan(x/2)]+2/[1+tan(x/2)]+C
=2(x/2)+2/[1+tan(x/2)]+C
=x+2/[1+tan(x/2)]+C
第三条:
∫dx/(3+cosx),u=tan(x/2),dx=2du/(1+u²)
=∫{1/[3+(1-u²)/(1+u²)]·2/(1+u²)}du
=∫du/(2+u²),令u=√2tanT,du=√2sec²TdT
=√2∫sec²T/(2+2tan²T) dT
=(√2/2)∫sec²T/sec²T dT
=(1/√2)T+C
=(1/√2)arctan(u/√2)
=(1/√2)arctan[(1/√2)tan(x/2)]+C
第四条:
∫dx/(1+sinx+cosx),u=tan(x/2),dx=2du/(1+u²)
=∫[1/[1+2u/(1+u²)+(1-u²)/(1+u²)]·2/(1+u²)]du
=∫du/(1+u)
=∫d(1+u)/(1+u)
=ln|1+u|+C
=ln|1+tan(x/2)|+C
∫1/(sinx+cosx)dx
=∫1/[√2sin(x+π/4)]dx
=√2/2∫1/sin(x+π/4)d(x+π/4)
令t=x+π/4则
上式=√2/2∫1/sint dt
=√2/2∫1/(2sint/2 cost/2) dt
=√2/2∫1/(tant/2 cos²t/2) dt/2
=√2/2∫1/(tant/2) d(tant/2)
=√2/2ln|tant/2|+C
故：
原式=√2/2ln|tan(x/2+π/8)|+C