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JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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∫0π/2f(sin⁡2x)sin⁡xdx=k∫0π/4f(cos⁡2x)cos⁡xdx\int\limits_{0}^{\pi / 2} f(\sin 2 x) \sin x d x=k \int\limits_{0}^{\pi / 4} f(\cos 2 x) \cos x d x where kk equals
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Correct option: C
2\sqrt{2}
Option C (2\sqrt{{2}}) is correct because after applying the properties of definite integrals and trigonometric identities, the value of k is found to be 2\sqrt{{2}}.

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