[toan 12] tich phân khó

N

nguyenbahiep1

[laTEX]I = \int_{0}^{\pi}x.cos^4x.sin^3x.dx \\ \\ x = \pi - u \Rightarrow dx = - du \\ \\ I = \int_{\pi}^{0}( \pi -u) .cos^4u.sin^3u.(-du) = \pi.\int_{0}^{\pi}.cos^4x.sin^3x.dx - \int_{0}^{\pi}.u .cos^4u.sin^3u.du \\ \\ I = \pi.\int_{0}^{\pi}.cos^4x.sin^3x.dx - I \Rightarrow 2.I = \pi.\int_{0}^{\pi}.cos^4x.sin^3x.dx[/laTEX]


đến đây dễ rồi nhé
 
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