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注:被积函数是偶函数,积分限关于原点对称,
故原式=2∫[0,+∞]e^(-x)cos²x dx
=∫[0,+∞]e^(-x)(1+cos2x)dx
=∫[0,+∞]e^(-x)dx+∫[0,+∞]e^(-x)cos2xdx
=-e^(-x)|[0,+∞] +∫[0,+∞]e^(-x)cos2xdx
=1+∫[0,+∞]e^(-x)cos2xdx①
=1 - ∫[0,+∞]cos2x d[e^(-x)]
=1 -e^(-x)cos2x|[0,+∞] - 2∫[0,+∞]e^(-x)sin2xdx
=1-(0-1)+2∫[0,+∞]sin2x d[e^(-x)]
=2+2e^(-x)sin2x|[0,+∞] - 4∫[0,+∞]e^(-x)cos2xdx
=2 -4∫[0,+∞]e^(-x)cos2xdx②
由①②得∫[0,+∞]e^(-x)cos2xdx=1/5
故原式=2-4∫[0,+∞]e^(-x)cos2xdx
=2- 4/5=6/5
故原式=2∫[0,+∞]e^(-x)cos²x dx
=∫[0,+∞]e^(-x)(1+cos2x)dx
=∫[0,+∞]e^(-x)dx+∫[0,+∞]e^(-x)cos2xdx
=-e^(-x)|[0,+∞] +∫[0,+∞]e^(-x)cos2xdx
=1+∫[0,+∞]e^(-x)cos2xdx①
=1 - ∫[0,+∞]cos2x d[e^(-x)]
=1 -e^(-x)cos2x|[0,+∞] - 2∫[0,+∞]e^(-x)sin2xdx
=1-(0-1)+2∫[0,+∞]sin2x d[e^(-x)]
=2+2e^(-x)sin2x|[0,+∞] - 4∫[0,+∞]e^(-x)cos2xdx
=2 -4∫[0,+∞]e^(-x)cos2xdx②
由①②得∫[0,+∞]e^(-x)cos2xdx=1/5
故原式=2-4∫[0,+∞]e^(-x)cos2xdx
=2- 4/5=6/5
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