Question:

Evaluate : \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \sqrt{2 \sin 2x}} \]

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For integrals involving square roots of trigonometric functions, consider substituting and using standard integrals for trigonometric powers.
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Solution and Explanation

We start by simplifying the expression inside the square root: \[ \sqrt{2 \sin 2x} = \sqrt{4 \sin x \cos x} = 2 \sqrt{\sin x \cos x} \] The integral becomes: \[ I = \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^3 x \cdot 2 \sqrt{\sin x \cos x}} = \frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{dx}{\cos^{\frac{5}{2}} x \sqrt{\sin x}} \] Now, we can apply a substitution to simplify further. Let \( u = \sin x \), then \( du = \cos x \, dx \). Changing the limits accordingly and performing the integration gives the final result.
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