∫(sinxcosx)/(sinx+cosx)dx=?
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∫ (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
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
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