Con tích phân này khó quá :((

L

langtuhoangminhtri

đặt [TEX]t=-x \Rightarrow dt=-dx[/TEX]
[TEX]x=\pi /2\rightarrow t=-\pi /2 , x=-\pi /2 \rightarrow t=\pi /2[/TEX]
nên [TEX]I= \int _{\pi/2}^{-\pi/2}\frac{costcos2te^t (-dt)}{e^t +1} = \int_{-\pi/2}^{\pi/2}\frac{cosxcos2x e^x dx}{e^x +1}[/TEX]
vậy [TEX]2I= \int_{-\pi/2}^{\pi/2}\frac{cosxcos2x e^x dx}{e^x +1} + \int_{-\pi/2}^{\pi/2}\frac{cosxcos2xdx}{e^x +1} = \int_{-\pi/2}^{\pi/2}cosxcos2xdx [/TEX]
ra rùi đó làm tiếp là ra KQ ngay thôi
 
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