Question

高数简单题！！在线等，急。

Answer 1

解：
令 x = tanu
原积分 = ∫ {cosu/ [ 1 + (sinu)^2 ]} du = ∫ d(sinu)/ [ 1 + (sinu)^2 ]
= arctan(sinu) + C
= arctan[ x/( 1 + x^2 )^(1/2) ] + C

追问

能不能详细一点啊？

追答

解：
令 x = tanu
原积分 = ∫ [du /(cosu^2 ] / {[ 2*(tanu)^2 + 1 ] * [ 1 + (tanu)^2 ] ^(1/2) }
= ∫ [du /(cosu^2 ] / {[ 2*(sinu/cosu)^2 + 1 ] * [ 1 /(cosu)^2 ] ^(1/2) }
= ∫ [du / {[ 2*(sinu)^2 + (cosu)^2 ] * [ 1 /(cosu) ] }
= ∫ [du / {[ ( (sinu)^2 + (cosu)^2 ) + (sinu)^2 ] * [ 1 /(cosu) ] }
= ∫ [du / {[ 1 + (sinu)^2 ] * [ 1 /(cosu) ] }
= ∫ [ cosu * du / {[ 1 + (sinu)^2 ]
= ∫ d(sinu)/ [ 1 + (sinu)^2 ]
= ∫ d [ arctan(sinu) ]
= arctan(sinu) + C ,
= arctan[ x/( 1 + x^2 )^(1/2) ] + C
这里， 因为 x = tanu = sinu/cosu
所以 x^2 = ( sinu/cosu )^2 = (sinu)^2 / (cosu)^2 = (sinu)^2 / [ 1 - (sinu)^2 ]
从而 x^2 / ( 1 + x^2 ) = (sinu)^2
两边开平方，有 sinu = [ x^2 / ( 1 + x^2 ) ]^(1/2) = x / ( 1 + x^2 )^(1/2)
所以 arctan(sinu) = arctan [ x / ( 1 + x^2 )^(1/2) ]