2024 ASRJC P2 Q1
2024 ASRJC P2 Q1
4 marks
Prelims
By using the substitution \(x=\cot \theta \), for \(0<\theta <\frac{\pi }{4}\), find \(\displaystyle\displaystyle \displaystyle\int \frac{1}{{{x}^{2}}\sqrt{1+{{x}^{2}}}}\,\mathrm{d}x\).[4]
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Answer:\(-\dfrac{\sqrt{1+x^2}}{x} + C\)
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