hept minh voi

N

nguyenhanhnt2012

sai chính tả...................................................................................................
 
N

nguyenbahiep1

\int_{bi^2/4}^{0}x*sincanx
hept minh voi cac men oj.thank nha.............................................


[TEX]\int_{0}^{\frac{\pi^2}{4}}.x.sin(\sqrt{x})dx \\ \sqrt{x} = u \Rightarrow x = u^2 \Rightarrow dx = 2udu \\ \int_{0}^{\frac{\pi}{2}}.2u.u^2.sinudu= \int_{0}^{\frac{\pi}{2}}.2u^3.sinu.du[/TEX]

đến đây tích phân từng phần 3 lần là ra đáp án

[TEX]\int_{0}^{\frac{\pi}{2}}.2u^3.sinu.du = -cosu.2u^3 + \int_{0}^{\frac{\pi}{2}}6u^2.cosu.du \\ 6u^2.sinu - \int_{0}^{\frac{\pi}{2}}12u.sinu.du \\ \frac{3.\pi^2}{2} + cosu.12u - \int_{0}^{\frac{\pi}{2}}12.cosu.du \\ \frac{3.\pi^2}{2} -12sinu = \frac{3.\pi^2}{2} - 12[/TEX]
 
Last edited by a moderator:
Top Bottom