题目
关于三角函数求不定积分的问题
$sinxcosx/(sinx+cosx)dx请数学牛人帮帮忙!
$sinxcosx/(sinx+cosx)dx请数学牛人帮帮忙!
提问时间：2020-12-06
答案
∫（sinxcosx）/（sinx+cosx）dx
=∫（2sinxcosx）/[2√2sin(x+π/4)]dx
=∫sin2x/[2√2sin(x+π/4)]dx
=∫-cos(2x+π/2)/[2√2sin(x+π/4)]dx
=∫[2sin^2(x+π/4)-1]/[2√2sin(x+π/4)]dx
=∫｛√2/2*sin(x+π/4)+√2/4[-1/sin(x+π/4)]｝dx
=∫√2/2*sin(x+π/4)dx+∫√2/4[-1/sin(x+π/4)]dx
=-√2/2*cos(x+π/4)+√2/4*cos(x+π/4)/sin(x+π/4)+C
=∫（2sinxcosx）/[2√2sin(x+π/4)]dx
=∫sin2x/[2√2sin(x+π/4)]dx
=∫-cos(2x+π/2)/[2√2sin(x+π/4)]dx
=∫[2sin^2(x+π/4)-1]/[2√2sin(x+π/4)]dx
=∫｛√2/2*sin(x+π/4)+√2/4[-1/sin(x+π/4)]｝dx
=∫√2/2*sin(x+π/4)dx+∫√2/4[-1/sin(x+π/4)]dx
=-√2/2*cos(x+π/4)+√2/4*cos(x+π/4)/sin(x+π/4)+C
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