Question

Find: \[ \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx. \]

Hint:

. \textbf{Part (b):} For integrals involving roots and trigonometric functions, use substitutions like \(\sin 2x = 2 \sin x \cos x\) to simplify the square root terms and reduce the integrand into an easily integrable form.

Answer 1

The integral to be evaluated is:\[I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{\sin 2x}} \, dx.\]1. Integrand Simplification: Apply the double angle identity \( \sin 2x = 2\sin x \cos x \). This yields: \[ \sqrt{\sin 2x} = \sqrt{2\sin x \cos x}. \] Substituting this into the integral gives: \[ I = \int_{\pi/6}^{\pi/3} \frac{\sin x + \cos x}{\sqrt{2\sin x \cos x}} \, dx. \]2. Further Simplification: Rewrite \( \sin x + \cos x \) using the identity \( \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right) \). The integral becomes: \[ I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)}{\sqrt{2} \sqrt{\sin x \cos x}} \, dx. \] Canceling the \( \sqrt{2} \) terms results in: \[ I = \int_{\pi/6}^{\pi/3} \frac{\sin\left(x + \frac{\pi}{4}\right)}{\sqrt{\sin x \cos x}} \, dx. \]3. Substitution: A substitution such as \( u = \sin x \), with \( du = \cos x \, dx \), can be considered, but the complete evaluation requires additional steps and transformations.Both integrals necessitate advanced integration techniques or numerical approximation methods for their solution. Computational tools may be required for practical results.\bigskip