Step 1: Simplify the denominator and numerator.
1. For the denominator \(9 + 16 \sin 2x\), use the identity \(\sin 2x = 2 \sin x \cos x\): \[ 9 + 16 \sin 2x = 9 + 16(2 \sin x \cos x) = 9 + 32 \sin x \cos x. \]
2. For the numerator \(\sin x + \cos x\), use the identity: \[ \sin x + \cos x = \sqrt{2} \sin\left(x + \frac{\pi}{4}\right). \] Thus, the integral becomes: \[ \int_{0}^{\frac{\pi}{4}} \frac{\sqrt{2} \sin\left(x + \frac{\pi}{4}\right)}{9 + 32 \sin x \cos x} \, dx. \]
Step 2: Substitution for simplification. Let \(\sin x = t\).
Then: \[ \cos x \, dx = dt. \] The limits of integration change as follows:
When \(x = 0\), \(\sin x = 0 \implies t = 0\),
When \(x = \frac{\pi}{4}\), \(\sin x = \frac{\sqrt{2}}{2} \implies t = \frac{\sqrt{2}}{2}\).
Using the substitution \(\sin x = t\), \(\cos x = \sqrt{1 - t^2}\), and \(\sin 2x = 2t\sqrt{1 - t^2}\), the integral becomes: \[ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{\sqrt{2} \cdot \sin\left(\arcsin t + \frac{\pi}{4}\right)}{9 + 32 \cdot t \sqrt{1 - t^2}} \cdot \frac{dt}{\sqrt{1 - t^2}}. \]
Step 3: Simplify the trigonometric terms. Using the identity \(\sin(a + b) = \sin a \cos b + \cos a \sin b\): \[ \sin\left(\arcsin t + \frac{\pi}{4}\right) = t \cdot \frac{\sqrt{2}}{2} + \sqrt{1 - t^2} \cdot \frac{\sqrt{2}}{2}. \]
Substitute this back: \[ \int_{0}^{\frac{\sqrt{2}}{2}} \frac{\sqrt{2} \left[t \cdot \frac{\sqrt{2}}{2} + \sqrt{1 - t^2} \cdot \frac{\sqrt{2}}{2}\right]}{9 + 32t\sqrt{1 - t^2}} \cdot \frac{dt}{\sqrt{1 - t^2}}. \]
Simplify further and evaluate this integral, which can be computed directly or using numerical methods. Final Answer: The exact evaluation of this integral is tedious and may involve further simplifications or computational techniques. The simplified form of the integrand allows easier evaluation using numerical methods.
A compound (A) with molecular formula $C_4H_9I$ which is a primary alkyl halide, reacts with alcoholic KOH to give compound (B). Compound (B) reacts with HI to give (C) which is an isomer of (A). When (A) reacts with Na metal in the presence of dry ether, it gives a compound (D), C8H18, which is different from the compound formed when n-butyl iodide reacts with sodium. Write the structures of A, (B), (C) and (D) when (A) reacts with alcoholic KOH.