∫(sinxcosx)⼀(sinx+

∫ (sinxcosx)/(sinx + cosx) dx=(1/2)(- cosx + sinx) - [1/(2√2)]ln|csc(x + π/4) - cot(x + π/4)| + C。C为积分常数。

解答过程如下：

∫ (sinxcosx)/(sinx + cosx) dx

= (1/2)∫ (2sinxcosx)/(sinx + cosx) dx

= (1/2)∫ [(1 + 2sinxcosx) - 1]/(sinx + cosx) dx

= (1/2)∫ (sin²x + 2sinxcosx + cos²x)/(sinx + cosx) dx - (1/2)∫ dx/(sinx + cosx)

= (1/2)∫ (sinx + cosx)²/(sinx + cosx) dx - (1/2)∫ dx/[√2sin(x + π/4)]

= (1/2)∫ (sinx + cosx) dx - [1/(2√2)]∫ csc(x + π/4) dx

= (1/2)(- cosx + sinx) - [1/(2√2)]ln|csc(x + π/4) - cot(x + π/4)| + C