2020_II_1_Pure

2020_II_1_Pure

Q1 - Solution.pdf (35.4 KB)

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This is so well explained

alternative solution (perhaps more long winded but a bit different nonetheless)

(i) we have the integral

\begin{align} \int\dfrac{dx}{x^{3/2}\sqrt{x-1}} \\ x=\dfrac{1}{1-u}\implies dx=\dfrac{du}{(1-u)^2} \\\therefore \int\dfrac{dx}{x^{3/2}\sqrt{x-1}} &=\int\dfrac{(1-u)^{-2}du}{\sqrt{\dfrac{u}{1-u}}(1-u)^{-3/2}} \\&=\int\dfrac{(1-u)^{-2}(1-u)^{3/2}\sqrt{(1-u)}du}{\sqrt{u}} \\&=\int\dfrac{1}{\sqrt{u}}du \\&=2\sqrt{u}+c \\&=2\sqrt{\dfrac{x-1}{x}}+c \end{align}

(ii) we now have the integral

\begin{align} \int\dfrac{dx}{(x-2)^{3/2}\sqrt{x+1}} \\\gamma=x-2\implies d\gamma=dx \\\therefore\int\dfrac{dx}{(x-2)^{3/2}\sqrt{x+1}} =\int\dfrac{d\gamma}{\gamma^{3/2}\sqrt{\gamma+3}} \\\alpha=\dfrac{1}{\gamma + 3} \implies d\alpha=-\dfrac{d\gamma}{(\gamma+3)^{2}} \iff d\gamma=-(\gamma+3)^{2} d\alpha=-\alpha^{-2} d\alpha \\\therefore\int\dfrac{dx}{\gamma^{3/2}\sqrt{\gamma+3}} &=\int\dfrac{-\alpha^{-2}d\alpha}{\left(\frac{1-3\alpha}{\alpha}\right)^{3/2}\alpha^{-1/2}} \\&=\int\dfrac{-\alpha^{2}d\alpha}{(1-3\alpha)^{3/2}\alpha^{2}} \\&=\int -(1-3\alpha)^{-3/2}d\alpha \\&=\dfrac{-2}{3} (1-3\alpha)^{-1/2}+c \\&=\dfrac{-2}{3} \sqrt{\dfrac{x+1}{x-2}}+c \end{align}

(iii) finally we have

\begin{align} \int_{2}^{\infty}\dfrac{dx}{(x-1)\sqrt{x-2}\sqrt{3x-2}} \\u=x-1 \implies du=dx \\\therefore \int_{2}^{\infty} \dfrac{dx}{(x-1)\sqrt{x-2}\sqrt{3x-2}} =\int_{1}^{\infty}\dfrac{du}{u \sqrt{u-1}\sqrt{3u+1}} \\w=\dfrac{1}{3u+1}\implies du=\dfrac{-1}{3} w^{-2}dw \\\therefore \int_{1/4}^{0}\dfrac{-\frac{1}{3}w^{-2}dw}{\left(\dfrac{1-w}{3w}\right)\sqrt{\dfrac{1-4w}{3w}}\left(\dfrac{1}{\sqrt{w}}\right)} &=\dfrac{1}{3}\int_{0}^{1/4}\dfrac{w^{-2}w^{2}(3)^{3/2}dw}{(1-w)\sqrt{1-4w}} \\&=\sqrt{3}\int_{0}^{1/4}\dfrac{dw}{(1-w)\sqrt{1-4w}} \\\beta=\dfrac{1}{1-4w} \implies dw=\dfrac{1}{4}\beta^{-2} \\\therefore \sqrt{3}\int_{0}^{1/4}\dfrac{dw}{(1-w)\sqrt{1-4w}} &=\dfrac{\sqrt{3}}{4} \int_{1}^{\infty}\dfrac{\beta^{-2}d\beta}{\left(\dfrac{3\beta+1}{4\beta}\right)\beta^{-1/2}} \\&=\dfrac{\sqrt{3}}{4} \int_{1}^{\infty}\dfrac{\beta^{-2}(4 {\beta})\sqrt{\beta}d \beta}{3 \beta +1} \\&=\sqrt{3}\int_{1}^{\infty}\dfrac{d\beta}{\sqrt{\beta}(3\beta+1)} \\\beta=\dfrac{1}{3}\tan^2\theta \implies d\beta=\dfrac{2}{3}\tan \theta \sec^{2}\theta d\theta \\\therefore \sqrt{3}\int_{1}^{\infty}\dfrac{d\beta}{\sqrt{\beta}(3\beta+1)} &=\sqrt{3}\int_{\pi/3}^{\pi/2}\dfrac{\frac{2}{3}\tan\theta \sec^{2}\theta d\theta}{\frac{1}{\sqrt{3}} \tan \theta (\tan^{2} \theta +1)} \\&=2\int_{\pi/3}^{\pi/2}\dfrac{\tan\theta \sec^{2}\theta d\theta}{\tan \theta \sec^{2} \theta} \\&=2\int_{\pi/3}^{\pi/2} 1 d \theta \\&=2(\pi/2-\pi/3) \\&=\dfrac{2\pi}{6} \\&=\dfrac{\pi}{3} \end{align}
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