If I1=∫0π/2f(sin2x) sinx dx and I2=∫0π/4f(cos2x) cosx dx then I1I2=

I1=∫0π/2f(sin2(π2−x)) sin(π2−x) dx 2I1=∫0π/2f (sin2x)(sinx+cosx)dx 2∫0π/2f(sin 2x) cos(x−π4)dx x−π4=t =2 ∫−π/4π/4f(cos2t) cost dt 2I1=22 ∫0π/4f(cos2t) cost dt 2I1=22 I2 I1I2=2

If I1=∫0π/2f(sin2x) sinx dx and I2=∫0π/4f(cos2x) cosx dx then I1I2=

I1=∫0π/2f(sin2(π2−x)) sin(π2−x) dx 2I1=∫0π/2f (sin2x)(sinx+cosx)dx 2∫0π/2f(sin 2x) cos(x−π4)dx x−π4=t =2 ∫−π/4π/4f(cos2t) cost dt 2I1=22 ∫0π/4f(cos2t) cost dt 2I1=22 I2 I1I2=2

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If $I_{1} = \int_{0}^{π / 2} f (\sin 2 x) \sin x d x$ and $I_{2} = \int_{0}^{π / 4} f (\cos 2 x) \cos x d x$ then $\frac{I_{1}}{I_{2}} =$

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$4 \sqrt{2}$

$2 \sqrt{2}$

$3 \sqrt{2}$

$\sqrt{2}$

answer is D.

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Detailed Solution

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