Hard +2.3 This AEA question requires sophisticated integration techniques across three parts: a non-trivial trigonometric substitution with careful limit handling, integration by parts with cotangent requiring multiple steps and algebraic manipulation, and a volume of revolution combining results from parts (a) and (b) with extensive algebraic simplification to reach a specific exact form. The multi-stage reasoning, technical manipulation demands, and need to synthesize earlier results place this well above typical A-level integration questions.
Use the substitution \(x = \sec\theta\) to show that
$$\int_{\sqrt{2}}^{2} \frac{1}{(x^2 - 1)^{\frac{3}{2}}} \, dx = \frac{\sqrt{6} - 2}{\sqrt{3}}$$
[5]
Use integration by parts to show that
$$\int \cos\theta \cot^2\theta \, d\theta = \frac{1}{2}[\ln|\cos\theta + \cot\theta| - \cos\theta \cot\theta] + c$$
[6]
% Figure shows a curve y = 1/(x^2-1)^(1/2) for x > 1, with shaded region R between x = sqrt(2) and x = 2
\includegraphics{figure_2}
Figure 2 shows a sketch of part of the curve with equation \(y = \frac{1}{(x^2 - 1)^{\frac{1}{2}}}\) for \(x > 1\)
The region \(R\), shown shaded in Figure 2, is bounded by the curve, the \(x\)-axis and the lines \(x = \sqrt{2}\) and \(x = 2\)
The region \(R\) is rotated through \(2\pi\) radians about the \(x\)-axis.
Show that the volume of the solid formed is
$$\pi \left[\frac{3}{8}\ln\left(\frac{1 + \sqrt{2}}{\sqrt{3}}\right) + \frac{7}{36} - \frac{\sqrt{2}}{8}\right]$$
[8]
(a) Use the substitution $x = \sec\theta$ to show that
$$\int_{\sqrt{2}}^{2} \frac{1}{(x^2 - 1)^{\frac{3}{2}}} \, dx = \frac{\sqrt{6} - 2}{\sqrt{3}}$$
[5]
(b) Use integration by parts to show that
$$\int \cos\theta \cot^2\theta \, d\theta = \frac{1}{2}[\ln|\cos\theta + \cot\theta| - \cos\theta \cot\theta] + c$$
[6]
% Figure shows a curve y = 1/(x^2-1)^(1/2) for x > 1, with shaded region R between x = sqrt(2) and x = 2
\includegraphics{figure_2}
Figure 2 shows a sketch of part of the curve with equation $y = \frac{1}{(x^2 - 1)^{\frac{1}{2}}}$ for $x > 1$
The region $R$, shown shaded in Figure 2, is bounded by the curve, the $x$-axis and the lines $x = \sqrt{2}$ and $x = 2$
The region $R$ is rotated through $2\pi$ radians about the $x$-axis.
(c) Show that the volume of the solid formed is
$$\pi \left[\frac{3}{8}\ln\left(\frac{1 + \sqrt{2}}{\sqrt{3}}\right) + \frac{7}{36} - \frac{\sqrt{2}}{8}\right]$$
[8]
\hfill \mbox{\textit{Edexcel AEA 2015 Q7 [19]}}