7.求一道 高数题
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设x=arctanu,则dx=du/(1+u^2),
∫sinxdx/(2sinx+cosx)
=∫udu/[(2u+1)(u^2+1)]
=(1/5)∫[(u+2)/(u^2+1)-2/(2u+1)]du
=(1/5)[(1/2)ln(u^2+1)+2arctanu-ln|u+1/2|]+c
=(1/5)[-ln|cosx|+2x-ln|tanx+1/2|]+c.
=(1/5)[2x-ln|sinx+(1/2)cosx|]+c.
∫sinxdx/(2sinx+cosx)
=∫udu/[(2u+1)(u^2+1)]
=(1/5)∫[(u+2)/(u^2+1)-2/(2u+1)]du
=(1/5)[(1/2)ln(u^2+1)+2arctanu-ln|u+1/2|]+c
=(1/5)[-ln|cosx|+2x-ln|tanx+1/2|]+c.
=(1/5)[2x-ln|sinx+(1/2)cosx|]+c.
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