Question

If \( \int_{\pi/4}^{3\pi/4} \frac{x \sin x}{1+3\cos 2x} \, dx = k \int_{\pi/4}^{3\pi/4} \frac{\sin x}{1+3\cos 2x} \, dx \), then \( \int_{0}^{k} \sin^{\pi/k}x \, dx = \)

• A. \( \frac{2}{3} \)
• B. \( \frac{\pi}{2} \)
• C. \( \frac{3\pi}{4} \)
• D. \( \frac{\pi}{4} \)

Hint:

The definite integral value of both \( \int_{0}^{\pi/2} \sin^2 x \, dx \) and \( \int_{0}^{\pi/2} \cos^2 x \, dx \) is always equal to exactly \( \frac{\pi}{4} \). This is a very common landmark value worth memorizing.

Answer 1

The Correct Option is D

Concept: According to the definite integral property (King’s Rule): \[ \int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a+b-x) \, dx \]

Step 1: Applying King's Rule to find parameter \( k \).
Here, the limits sum up to \( a + b = \frac{\pi}{4} + \frac{3\pi}{4} = \pi \). Replace \( x \) with \( \pi - x \): \[ I = \int_{\pi/4}^{3\pi/4} \frac{(\pi-x)\sin(\pi-x)}{1+3\cos 2(\pi-x)} \, dx = \int_{\pi/4}^{3\pi/4} \frac{(\pi-x)\sin x}{1+3\cos 2x} \, dx \] Adding this new integral expression to the original one: \[ 2I = \int_{\pi/4}^{3\pi/4} \frac{[x + (\pi-x)]\sin x}{1+3\cos 2x} \, dx = \pi \int_{\pi/4}^{3\pi/4} \frac{\sin x}{1+3\cos 2x} \, dx \] Dividing by 2: \[ I = \frac{\pi}{2} \int_{\pi/4}^{3\pi/4} \frac{\sin x}{1+3\cos 2x} \, dx \] Comparing with the given form, we find: \[ k = \frac{\pi}{2} \]

Step 2: Evaluating the target integral.
Substitute \( k = \frac{\pi}{2} \) into the second integral expression: \[ \int_{0}^{\pi/2} \sin^{\pi/(\pi/2)}x \, dx = \int_{0}^{\pi/2} \sin^{2}x \, dx \]

Step 3: Computing the final value using symmetric properties.
Using the identity \( \int_{0}^{\pi/2} \sin^2 x \, dx = \frac{\pi}{4} \), let's ensure standard definite integral calculations are fully aligned: \[ \int_{0}^{\pi/2} \sin^2 x \, dx = \int_{0}^{\pi/2} \frac{1 - \cos 2x}{2} \, dx = \left[ \frac{x}{2} - \frac{\sin 2x}{4} \right]_{0}^{\pi/2} = \frac{\pi}{4} \]