Integration Techniques: Algebraic sub; trig sub — JPJC 2025 H2 Math Prelim Paper 1

(a)(i) [1 mark]

\[\begin{aligned} \int\frac{x}{\sqrt{25-x^{2}}}\,\mathrm{d}x &= -\tfrac{1}{2}\int -2x(25-x^{2})^{-1/2}\,\mathrm{d}x\\ &= -\sqrt{25-x^{2}}+C \end{aligned}\]

(a)(ii) [4 marks]

\[\begin{aligned} \int_{\alpha}^{4}\left|\frac{x}{\sqrt{25-x^{2}}}\right|\mathrm{d}x &= 3\\ -\int_{\alpha}^{0}\frac{x}{\sqrt{25-x^{2}}}\,\mathrm{d}x + \int_{0}^{4}\frac{x}{\sqrt{25-x^{2}}}\,\mathrm{d}x &= 3\\ -\left[-\sqrt{25-x^{2}}\right]_{\alpha}^{0} + \left[-\sqrt{25-x^{2}}\right]_{0}^{4} &= 3\\ \left[\sqrt{25-x^{2}}\right]_{\alpha}^{0} - \left[\sqrt{25-x^{2}}\right]_{0}^{4} &= 3\\ 5-\sqrt{25-\alpha^{2}}-[3-5] &= 3\\ \sqrt{25-\alpha^{2}} &= 4\\ \alpha^{2} &= 9\\ \alpha &= \pm 3 \end{aligned}\]

Since \(\alpha < 0\), \(\alpha = -3\).

(b) [5 marks]

Let \(x = 4\tan\theta\), so \(\dfrac{\mathrm{d}x}{\mathrm{d}\theta} = 4\sec^{2}\theta\).

When \(x = 4\): \(\tan\theta = 1 \;\Rightarrow\; \theta = \dfrac{\pi}{4}\).

When \(x = 0\): \(\tan\theta = 0 \;\Rightarrow\; \theta = 0\).

\[\begin{aligned} \int_{0}^{4}\sqrt{\frac{x^{2}}{16+x^{2}}}\,\mathrm{d}x &= \int_{0}^{\pi/4}\sqrt{\frac{16\tan^{2}\theta}{16+16\tan^{2}\theta}}\;4\sec^{2}\theta\,\mathrm{d}\theta\\ &= \int_{0}^{\pi/4}\sqrt{\frac{\tan^{2}\theta}{\sec^{2}\theta}}\;4\sec^{2}\theta\,\mathrm{d}\theta\\ &= 4\int_{0}^{\pi/4}\frac{\tan\theta}{\sec\theta}\sec^{2}\theta\,\mathrm{d}\theta\\ &= 4\int_{0}^{\pi/4}\tan\theta\sec\theta\,\mathrm{d}\theta\\ &= 4\Bigl[\sec\theta\Bigr]_{0}^{\pi/4}\\ &= 4\!\left[\sec\tfrac{\pi}{4}-\sec 0\right]\\ &= 4(\sqrt{2}-1) \end{aligned}\]