<ol class="abc-list"><li><div class="list-item-content"><span class="list-item-text">Use the substitution \(x=\sin^2 \theta\), where \(0 \leq \theta \leq \frac{\pi}{2}\), to find \(\int_0^{\frac{1}{2}} \sqrt{\frac{16x}{1-x}} \, \mathrm{d}x\) exactly.</span><span class="inline-marks">[4]</span></div></li><li><div class="list-item-content"><span class="list-item-text">Find \(\int_a^b \frac{|x-1|}{\sqrt{\frac{1}{2}x^2 - x + 1}} \, \mathrm{d}x\) in terms of \(a\) and \(b\), where \(a &lt; 1 &lt; b\).</span><span class="inline-marks">[4]</span></div></li></ol>

Use the substitution \(x=\sin^2 \theta\), where \(0 \leq \theta \leq \frac{\pi}{2}\), to find \(\displaystyle \int_0^{\frac{1}{2}} \sqrt{\frac{16x}{1-x}} \, \mathrm{d}x\) exactly.

Find \(\displaystyle \int_a^b \frac{|x-1|}{\sqrt{\frac{1}{2}x^2 - x + 1}} \, \mathrm{d}x\) in terms of \(a\) and \(b\), where \(a < 1 < b\).

2025 NYJC P1 Q7

2025 NYJC P1 Q7