Question: Evaluate: $\int_{0}^{\pi} \frac{x}{1 + \sin x} dx$

Step 1 (Formula / Governing Law & Given Data): [1 Mark]

Let $I = \int_{0}^{\pi} \frac{x}{1 + \sin x} dx$ --- (i)

We apply the King's Property (Definite Integral Property):

$$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$$

Here, $a = 0$ and $b = \pi$, so we replace $x$ with $(\pi - x)$.

Step 2 (Step-by-Step Mathematical Working): [2 Marks]

Applying the property to equation (i):

$$I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin(\pi - x)} dx$$

Since $\sin(\pi - x) = \sin x$, we have:

$$I = \int_{0}^{\pi} \frac{\pi - x}{1 + \sin x} dx$$ --- (ii)

Adding (i) and (ii):

$$2I = \int_{0}^{\pi} \frac{x + \pi - x}{1 + \sin x} dx = \pi \int_{0}^{\pi} \frac{1}{1 + \sin x} dx$$

Rationalizing the denominator:

$$2I = \pi \int_{0}^{\pi} \frac{1 - \sin x}{(1 + \sin x)(1 - \sin x)} dx = \pi \int_{0}^{\pi} \frac{1 - \sin x}{\cos^2 x} dx$$

$$2I = \pi \int_{0}^{\pi} (\sec^2 x - \sec x \tan x) dx$$

Integrating the terms:

$$2I = \pi [\tan x - \sec x]_{0}^{\pi}$$

Step 3 (Final Concluded Answer): [1 Mark]

Evaluating the limits:

$$2I = \pi [(\tan \pi - \sec \pi) - (\tan 0 - \sec 0)]$$

$$2I = \pi [(0 - (-1)) - (0 - 1)]$$

$$2I = \pi [1 + 1] = 2\pi$$

$$I = \pi$$

Final Answer: The value of the integral is $\pi$.

💡 EL10 Concept Insight: Think of the "King's Property" as a mirror trick; by replacing $x$ with $(\pi - x)$, we create a second version of the integral where the messy $x$ in the numerator cancels out, leaving us with a much simpler fraction that is easy to solve!