Use the substitution $x=\sin^2 \theta$, where $0 \leq \theta \leq \frac{\pi}{2}$, to find \

Question

Use the substitution $x=\sin^2 \theta$, where $0 \leq \theta \leq \frac{\pi}{2}$, to find $\displaystyle \displaystyle\int_0^{\frac{1}{2}} \sqrt{\frac{16x}{1-x}} \, \mathrm{d}x$ exactly.[4] Find $\displaystyle \displaystyle\int_a^b \frac{|x-1|}{\sqrt{\frac{1}{2}x^2 - x + 1}} \, \mathrm{d}x$ in terms of $a$ and $b$, where $a < 1 < b$.[4]

Accepted Answer

(a) $\pi-2$ (b) $2\sqrt{\dfrac12a^2-a+1}+2\sqrt{\dfrac12b^2-b+1}-2\sqrt2$

2025 NYJC P1 Q7

2025 NYJC P1 Q7

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