Let f(x) be an even function, I1=∫0π2f(cos2x)cosxdx,I2=∫0π4f(sin2x)cosxdx . Then I1I2=

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Let $f (x)$ be an even function, $I_{1} = \int_{0}^{\frac{π}{2}} f (\cos 2 x) \cos x d x, I_{2} = \int_{0}^{\frac{π}{4}} f (\sin 2 x) \cos x d x$ . Then $\frac{I_{1}}{I_{2}} =$

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$\frac{1}{2}$

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

answer is D.

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$I_{1} = \int_{0}^{\frac{π}{4}} f (\cos 2 x) \cos x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} f (\cos 2 x) \cos x d x$

Setting $x = \frac{π}{4} - t$ in the first integral and $x = \frac{π}{4} + t$ in the second integral,

$I_{1} = \int_{0}^{\frac{π}{4}} f (\sin 2 t) \cos (\frac{π}{4} - t) d t + \int_{0}^{\frac{π}{4}} f (\sin 2 t) \cos (\frac{π}{4} + t) d t$

$= \int_{0}^{\frac{π}{4}} f (\sin 2 t) [\cos (\frac{π}{4} - t) + \cos (\frac{π}{4} + t)] d t$

$= \sqrt{2} \int_{0}^{\frac{π}{4}} f (\sin 2 t) \cos t d t = \sqrt{2} I_{2}$

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