高等数学之不定积分求解设f(x)=cosxsinx/(cosx+sinx),求f(x)的不定积分这道题很难的,
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高等数学之不定积分求解设f(x)=cosxsinx/(cosx+sinx),求f(x)的不定积分这道题很难的,
高等数学之不定积分求解
设f(x)=cosxsinx/(cosx+sinx),求f(x)的不定积分
这道题很难的,
高等数学之不定积分求解设f(x)=cosxsinx/(cosx+sinx),求f(x)的不定积分这道题很难的,
好题
cosxsinx/(cosx+sinx),cosx+sinx=√2(√2/2cosx+√2/2sinx)=√2cos(x-π/4) cosx=cos(x-π/4+π/4)=√2/2[cos(x-π/4)-sin(x-π/4)] sinx=sin(x-π/4+π/4)=√2/2[cos(x-π/4)+sin(x-π/4)] cosxsinx=(1/2)[(cos(x-π/4))^2-(sin(x-π/4))^2] 积分为∫(1/2)[(cos(x-π/4))^2-(sin(x-π/4))^2]dx/√2cos(x-π/4)=(1/2√2)∫[(cos(x-π/4))^2-(sin(x-π/4))^2]d(x-π/4)/cos(x-π/4) 设u=x-π/4,积分化为(1/2√2)∫[(cosu)^2-(sinu)^2]du/cosu=(1/2√2)∫[cosu-(sinu)^2/cosu]du=(1/2√2)∫[cosu-(sinu)^2*cosu/(cosu)^2]du=(1/2√2)∫cosudu-(1/2√2)∫[(sinu)^2*cosu/(cosu)^2]du=(1/2√2)sinu-(1/2√2)∫(sinu)^2d(sinu)/[1-(sinu)^2] 设sinu=v,则积分又化为(1/2√2)sinu-(1/2√2)∫v^2dv/[1-v^2]=(1/2√2)sinu+(1/2√2)∫[(v^2-1+1)/(v^-1)]dv=(1/2√2)sinu+(1/2√2)∫[1+1/(v^2-1)]dv=(1/2√2)sinu+(1/2√2)[v+(1/2)ln|(v-1)/(v+1)|] 接下来 ,将v全部换回得(1/2√2)sinu+(1/2√2)[sinu+(1/2)ln|(sinu-1)/(sinu+1)|]=(1/2√2)[2sinu+(1/2)ln|(sinu-1)/(sinu+1)|] 再将u换回,得积分=(1/2√2)[2sinu+(1/2)ln|(sinu-1)/(sinu+1)|]=(1/2√2)[2sin(x-π/4)+(1/2)ln|(sin(x-π/4)-1)/(sin(x-π/4)+1)|]+C
我不是他舅&我不是ta舅
不是很难嘛
∫cosxsinx/(cosx+sinx) dx
=√2/4*∫sin2x/sin(x+π/4) dx
作换元 x+π/4=t
=-√2/4*∫cos2t/sint dt
∫cos2t/sint dt
=∫1/sint dt- 2∫sin^2t/sint dt
=ln|tant/2|+2cost+C
原积分
=-√2/4*ln|tant/2|-√2/2*cost+C
代入t即可