\int_{}^{} cos2x.(cosx^2006+sinx^2006).dx..can tu 0 den pi/2
\[\begin{array}{l}
I = \int {\cos 2x\left( {{{\cos }^{2006}}x + {{\sin }^{2006}}x} \right)dx = \int {\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\left( {{{\cos }^{2006}}x - {{\sin }^{2006}}x} \right)dx} } \\
= \int {{{\cos }^2}x\left( {{{\cos }^{2006}}x + {{\sin }^{2006}}x} \right)dx - \int {{{\sin }^2}x\left( {{{\cos }^{2006}}x - {{\sin }^{2006}}x} \right)dx} } = {I_1} - {I_2}\\
t = \frac{\pi }{2} - x \to dx = - dt\\
\to x = \frac{\pi }{2} - t\\
\left\{ \begin{array}{l}
x = 0 \to t = \frac{\pi }{2}\\
x = \frac{\pi }{2} \to t = 0
\end{array} \right.\\
\to {I_1} = - \int_{\frac{\pi }{2}}^0 {{{\cos }^2}\left( {\frac{\pi }{2} - t} \right)\left( {{{\cos }^{2006}}\left( {\frac{\pi }{2} - t} \right) + {{\sin }^{2006}}\left( {\frac{\pi }{2} - t} \right)} \right)} dt = \int_0^{\frac{\pi }{2}} {{{\sin }^2}t\left( {{{\cos }^{2006}}t + {{\sin }^{2006}}t} \right)dt} = {I_2}\\
\to I = {I_1} - {I_2} = 0
\end{array}\]
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