神马作文网如何求∫cosx/(sinx+cosx)dx,请知道的朋友帮帮忙,
来源：学生作业帮 编辑：神马作文网作业帮 分类：数学作业 时间：2024/11/17 17:02:32
如何求∫cosx/(sinx+cosx)dx,请知道的朋友帮帮忙,
解法一：设tan(x/2)=t
则x=2arctant,sinx=2t/(1+t²),cosx=(1-t²)/(1+t²),dx=2dt/(1+t²)
故∫cosx/(sinx+cosx)dx
=∫[(1-t²)/(1+t²)]*[2dt/(1+t²)]/[2t/(1+t²)+(1-t²)/(1+t²)]
=∫2(1-t²)dt/[(1+t²)(1+2t-t²)]
=∫[(1/2)/(t-1-√2)+(1/2)/(t-1+√2)+(1-t)/(1+t²)]dt
=(1/2)[ln│(t-1-√2)(t-1+√2)│]+arctant-(1/2)ln(1+t²)+C (C是积分常数)
=(1/2)ln│(t²-2t-1)/(1+t²)│+arctant+C
=(1/2)ln│(1-t²)/(1+t²)+(2t)/(1+t²)│+arctant+C
=(1/2)ln│cosx+sinx│+x/2+C
=(ln│cosx+sinx│+x)/2+C；
解法二：∫cosx/(sinx+cosx)dx
=(1/2)∫2cosx/(sinx+cosx)dx
=(1/2)∫(sinx+cosx+cosx-sinx)/(sinx+cosx)dx
=(1/2)∫[1+(cosx-sinx)/(sinx+cosx)]dx
=[∫dx+∫d(sinx+cosx)/(sinx+cosx)]/2
=(x+ln│sinx+cosx│)/2+C (C是积分常数).
则x=2arctant,sinx=2t/(1+t²),cosx=(1-t²)/(1+t²),dx=2dt/(1+t²)
故∫cosx/(sinx+cosx)dx
=∫[(1-t²)/(1+t²)]*[2dt/(1+t²)]/[2t/(1+t²)+(1-t²)/(1+t²)]
=∫2(1-t²)dt/[(1+t²)(1+2t-t²)]
=∫[(1/2)/(t-1-√2)+(1/2)/(t-1+√2)+(1-t)/(1+t²)]dt
=(1/2)[ln│(t-1-√2)(t-1+√2)│]+arctant-(1/2)ln(1+t²)+C (C是积分常数)
=(1/2)ln│(t²-2t-1)/(1+t²)│+arctant+C
=(1/2)ln│(1-t²)/(1+t²)+(2t)/(1+t²)│+arctant+C
=(1/2)ln│cosx+sinx│+x/2+C
=(ln│cosx+sinx│+x)/2+C；
解法二：∫cosx/(sinx+cosx)dx
=(1/2)∫2cosx/(sinx+cosx)dx
=(1/2)∫(sinx+cosx+cosx-sinx)/(sinx+cosx)dx
=(1/2)∫[1+(cosx-sinx)/(sinx+cosx)]dx
=[∫dx+∫d(sinx+cosx)/(sinx+cosx)]/2
=(x+ln│sinx+cosx│)/2+C (C是积分常数).
如何求∫cosx/(sinx+cosx)dx的不定微分 ,请知道的朋友帮帮忙 ,
如何求∫cosx/(sinx+cosx)dx,请知道的朋友帮帮忙,
求不定积分：∫ cosx/(sinx+cosx) dx
求∫cosx/(sinx+cosx)dx的不定积分
∫cosx / (cosx+sinx)dx
求(sinx/(cosx+sinx))dx的积分
∫ sinx+cosx/(sinx-cosx)^1/3 dx 求不定积分
求积分：∫ sinx*sinx/(1+cosx*cosx)dx
求∫sinx dx/(sinx+cosx)的积分,x/2-ln|sinx+cosx|+c
∫dx/（sinx+cosx)
∫(sinx-cosx)dx
∫sinx/(sinx-cosx)dx