Find $\displaystyle\displaystyle \displaystyle\int{{{\cot }^{2}} \; 2x\,}\mathrm{d}x$. [2] Write d

Question

Find $\displaystyle\displaystyle \displaystyle\int{{{\cot }^{2}} \; 2x\,}\mathrm{d}x$. [2] Write down the constants $A$ and $B$ such that, for all values of $x$, $-2x-1=A\left( -2x-2 \right)+B$. Hence, find $\displaystyle\displaystyle \displaystyle\int{\frac{-2x-1}{\sqrt{4-2x-{{x}^{2}}}}\,\mathrm{d}x}$.[4]

Accepted Answer

(a) $-\frac{1}{2}\cot 2x-x+C$ (b) $2\sqrt{4-2x-x^2} + \sin^{-1}\left(\frac{x+1}{\sqrt{5}}\right) + C$

2023 ASRJC Promo Q4

2023 ASRJC Promo Q4

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