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 $I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{1}{1 + \sqrt{\cot x}} \, dx$.

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

Rewrite the integrand: $f(x) = \frac{1}{1 + \frac{\sqrt{\cos x}}{\sqrt{\sin x}}} = \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}}$.

Applying the property where $a+b = \frac{\pi}{12} + \frac{5\pi}{12} = \frac{6\pi}{12} = \frac{\pi}{2}$:

$$ I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\sin(\frac{\pi}{2}-x)}}{\sqrt{\sin(\frac{\pi}{2}-x)} + \sqrt{\cos(\frac{\pi}{2}-x)}} \, dx = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \frac{\sqrt{\cos x}}{\sqrt{\cos x} + \sqrt{\sin x}} \, dx $$

Adding the two expressions for $I$:

$$ 2I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} \left( \frac{\sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} + \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} \right) \, dx $$ $$ 2I = \int_{\frac{\pi}{12}}^{\frac{5\pi}{12}} 1 \, dx = [x]_{\frac{\pi}{12}}^{\frac{5\pi}{12}} = \frac{5\pi}{12} - \frac{\pi}{12} = \frac{4\pi}{12} = \frac{\pi}{3} $$

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

$$ I = \frac{\pi}{6} $$

💡 EL10 Concept Insight: Imagine you are walking a path and the scenery looks exactly the same in reverse; by adding the "forward" path and the "backward" path together, the complicated parts cancel out, leaving you with a simple straight line to measure!

Solution (b)

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

We use the definition of the modulus function: $|x| = x$ if $x \ge 0$ and $|x| = -x$ if $x < 0$.

$$ \int_{-\frac{\pi}{6}}^{\frac{\pi}{2}} (\sin |x| + \cos |x|) \, dx = \int_{-\frac{\pi}{6}}^{0} (\sin(-x) + \cos(-x)) \, dx + \int_{0}^{\frac{\pi}{2}} (\sin x + \cos x) \, dx $$

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

Using $\sin(-x) = -\sin x$ and $\cos(-x) = \cos x$: