第1个回答 推荐于2018-06-18
设t=tanx/2,x=2arctant,dx=2dt/(1+t^2)
sinx=2t/(1+t^2),cosx=(1-t^2)/(1+t^2)
所以∫dx/(2sinx-cosx+5)=∫2dt/[2(2t)-(1-t^2)+5(1+t^2)]=∫dt/(3t^2+2t+2)=1/3∫dt/[(t+1/3)^2+5/9]
=3/5∫dt/[9/5(t+1/3)^2+1]=1/√5∫3/√5dt/[(3/√5(t+1/3))^2+1]=1/√5arctan(3/√5(t+1/3))+C
=1/√5arctan(3/√5(tanx/2+1/3))+C本回答被网友采纳
sinx=2t/(1+t^2),cosx=(1-t^2)/(1+t^2)
所以∫dx/(2sinx-cosx+5)=∫2dt/[2(2t)-(1-t^2)+5(1+t^2)]=∫dt/(3t^2+2t+2)=1/3∫dt/[(t+1/3)^2+5/9]
=3/5∫dt/[9/5(t+1/3)^2+1]=1/√5∫3/√5dt/[(3/√5(t+1/3))^2+1]=1/√5arctan(3/√5(t+1/3))+C
=1/√5arctan(3/√5(tanx/2+1/3))+C本回答被网友采纳
第2个回答 2017-06-18
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