Integration Techniques: Trig sub \(x=\sin^2\!\theta\); modulus — NYJC 2025 H2 Math Prelim Paper 1

(a) \(x = \sin^2\theta\), \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 2\sin\theta\cos\theta\).

When \(x=0\), \(\theta=0\). When \(x=\dfrac{1}{2}\), \(\theta=\dfrac{\pi}{4}\).

\[\begin{aligned} \int_0^{\frac{1}{2}}\sqrt{\frac{16x}{1-x}}\;\mathrm{d}x &= \int_0^{\frac{\pi}{4}}\sqrt{\frac{16\sin^2\theta}{1-\sin^2\theta}} \cdot 2\sin\theta\cos\theta\;\mathrm{d}\theta\\ &= \int_0^{\frac{\pi}{4}}\frac{4\sin\theta}{\cos\theta}\cdot 2\sin\theta\cos\theta\;\mathrm{d}\theta \quad\left(\because 0\le\theta\le\tfrac{\pi}{2},\;\sqrt{\cos^2\theta}=\cos\theta\right)\\ &= \int_0^{\frac{\pi}{4}} 8\sin^2\theta\;\mathrm{d}\theta\\ &= 4\int_0^{\frac{\pi}{4}} 1-\cos 2\theta\;\mathrm{d}\theta\\ &= \left[4\theta - 2\sin 2\theta\right]_0^{\frac{\pi}{4}}\\ &= \pi - 2 \end{aligned}\]

(b) \[\begin{aligned} \int_a^b \frac{|x-1|}{\sqrt{\frac{1}{2}x^2-x+1}}\;\mathrm{d}x &= \int_a^1 \frac{-(x-1)}{\sqrt{\frac{1}{2}x^2-x+1}}\;\mathrm{d}x + \int_1^b \frac{(x-1)}{\sqrt{\frac{1}{2}x^2-x+1}}\;\mathrm{d}x\\ &= -\left[\frac{\left(\frac{1}{2}x^2-x+1\right)^{\frac{1}{2}}}{\frac{1}{2}}\right]_a^1 + \left[\frac{\left(\frac{1}{2}x^2-x+1\right)^{\frac{1}{2}}}{\frac{1}{2}}\right]_1^b\\ &= -\left[2\!\left(\tfrac{1}{2}x^2-x+1\right)^{\!\frac{1}{2}}\right]_a^1 + \left[2\!\left(\tfrac{1}{2}x^2-x+1\right)^{\!\frac{1}{2}}\right]_1^b\\ &= -\!\left[2\!\left(\tfrac{1}{2}\right)^{\!\frac{1}{2}} - 2\!\left(\tfrac{1}{2}a^2-a+1\right)^{\!\frac{1}{2}}\right] + \left[2\!\left(\tfrac{1}{2}b^2-b+1\right)^{\!\frac{1}{2}} - 2\!\left(\tfrac{1}{2}\right)^{\!\frac{1}{2}}\right]\\ &= -2\sqrt{2} + 2\!\left(\tfrac{1}{2}a^2-a+1\right)^{\!\frac{1}{2}} + 2\!\left(\tfrac{1}{2}b^2-b+1\right)^{\!\frac{1}{2}} \end{aligned}\]