Solution (a)

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

We use the King's Property of Definite Integrals:

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

Let the given integral be:

$$I = \int_{0}^{\pi} \frac{e^{\cos x}}{e^{\cos x} + e^{-\cos x}} \, dx \quad \dots (1)$$

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

Applying the property where $a=0$ and $b=\pi$, we replace $x$ with $(\pi - x)$:

$$I = \int_{0}^{\pi} \frac{e^{\cos(\pi - x)}}{e^{\cos(\pi - x)} + e^{-\cos(\pi - x)}} \, dx$$

Since $\cos(\pi - x) = -\cos x$, the integral becomes:

$$I = \int_{0}^{\pi} \frac{e^{-\cos x}}{e^{-\cos x} + e^{\cos x}} \, dx \quad \dots (2)$$

Adding equations (1) and (2):

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

Step 3 (Final Concluded Answer): [0.5 Mark]

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

💡 EL10 Concept Insight: Imagine you have two identical paths that add up to the same total; because the function is symmetric around the midpoint, the two versions of the integral are actually the same, so you just calculate the area of a simple rectangle and divide by two!

Solution (b)

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

We use the method of Partial Fractions. Let:

$$\frac{2x+1}{(x+1)^2 (x-1)} = \frac{A}{x+1} + \frac{B}{(x+1)^2} + \frac{C}{x-1}$$

Multiplying by the denominator:

$$2x+1 = A(x+1)(x-1) + B(x-1) + C(x+1)^2$$

Step 2 (Step-by-Step Mathematical Working): [1.5 Marks]

• Set $x = 1$: $2(1)+1 = C(1+1)^2 \implies 3 = 4C \implies C = \frac{3}{4}$
• Set $x = -1$: $2(-1)+1 = B(-1-1) \implies -1 = -2B \implies B = \frac{1}{2}$
• Compare coefficients of $x^2$: $0 = A + C \implies A = -C = -\frac{3}{4}$

Substituting back into the integral:

$$\int \left( \frac{-3/4}{x+1} + \frac{1/2}{(x+1)^2} + \frac{3/4}{x-1} \right) \, dx$$

Step 3 (