Tính tích phân sau: ∫(sin^2005 x+x^2005)dx

Question

Tính tích phân sau:

\(\int _{-\pi }^{\pi }(\sin ^{2005} x+x^{2005} ){\rm d}x \)

Answer 1

\(I =\int _{-\pi }^{\pi }(\sin ^{2005} x+x^{2005} )dx\)

 Đặt \(t=-x\). Khi đó:
\(\begin{array}{l} {I_{2} =\int _{\pi }^{-\pi }\left[\sin ^{2005} \left(-t\right)+\left(-t\right)^{2005} \right]d\left(-t\right) =\int _{\pi }^{-\pi }(\sin ^{2005} t+t^{2005} )dt } \\ {=-\int _{-\pi }^{\pi }(\sin ^{2005} x+x^{2005} )dx =-I_{2} } \end{array} \)
Do đó \(I=\int _{-\pi }^{\pi }(\sin ^{2005} x+x^{2005} )dx =0\)