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Tính: [tex] \int _{o}^{\pi}\frac{x.sin x dx}{1+cos^2 x}[/tex]

[TEX]x=\pi-t[/TEX]

[TEX]I=\int _{o}^{\pi}\frac{(\pi-t).sin t dt}{1+cos^2 t}[/TEX]

[TEX]\Rightarrow I=\frac{\pi}{2}\int _{o}^{\pi}\frac{sin x dx}{1+cos^2 x}[/TEX]

[TEX]cosx=tanu\righ -sin xdx=(tan^2u+1)du[/TEX]

[TEX]\Rightarrow I=-\frac{\pi}{2}\int_{\frac{\pi}{4}}^{-\frac{\pi}{4}}\frac{(tan^2u+1)du}{tan^2u+1}=\frac{\pi^2}{4}[/TEX]
 
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