<p>In this question you may use expansions from the List of Formulae (MF26).</p>
<ol class="i-list"><li><div class="list-item-content"><span class="list-item-text">Find the Maclurin expansion of $\ln (\cos 2x)$ in ascending powers of $x$, up to and including the term ${{x}^{6}}$. State any value(s) of $x$ in the domain $0\le x\le \frac{1}{4}\pi $ for which the expansion is not valid.</span></div><p style="text-align: right;">[6]</p></li><li><div class="list-item-content"><span class="list-item-text">Use your expansion from part (i) and integration to find an approximate expression for $\displaystyle\int{\frac{\ln (\cos 2x)}{{{x}^{2}}}\mathrm{d}x}$. Hence find an approximate value for $\displaystyle\int\nolimits_{0}^{0.5}{\frac{\ln (\cos 2x)}{{{x}^{2}}}\mathrm{d}x}$, giving your answer to $4$ decimal places.</span></div><p style="text-align: right;">[3]</p></li><li><div class="list-item-content"><span class="list-item-text">Use your graphing calculator to find a second approximate value for $\displaystyle\int\nolimits_{0}^{0.5}{\frac{\ln (\cos 2x)}{{{x}^{2}}}\mathrm{d}x}$, giving your answer to $4$decimal places.</span></div><p style="text-align: right;">[1]</p></li></ol>

In this question you may use expansions from the List of Formulae (MF26).

Find the Maclurin expansion of $\ln (\cos 2x)$ in ascending powers of $x$, up to and including the term ${{x}^{6}}$. State any value(s) of $x$ in the domain $0\le x\le \frac{1}{4}\pi $ for which the expansion is not valid.

Use your expansion from part (i) and integration to find an approximate expression for $\displaystyle\displaystyle \displaystyle\int{\frac{\ln (\cos 2x)}{{{x}^{2}}}\mathrm{d}x}$. Hence find an approximate value for $\displaystyle\displaystyle \displaystyle\int\nolimits_{0}^{0.5}{\frac{\ln (\cos 2x)}{{{x}^{2}}}\mathrm{d}x}$, giving your answer to $4$ decimal places.

Use your graphing calculator to find a second approximate value for $\displaystyle\displaystyle \displaystyle\int\nolimits_{0}^{0.5}{\frac{\ln (\cos 2x)}{{{x}^{2}}}\mathrm{d}x}$, giving your answer to $4$decimal places.

Maclaurin Series and Power Series - General

Maclaurin Series and Power Series

International Baccalaureate Higher Level Mathematics

IB HL Math

N2018 P2 Q4

$\ln(\cos 2x) = \ln\left[1-\dfrac{(2x)^{2}}{2!}+\dfrac{(2x)^{4}}{4!}-\dfrac{(2x)^{6}}{6!}+\cdots\right]$

$= \ln\left[1-2x^{2}+\dfrac{2}{3}x^{4}-\dfrac{4}{45}x^{6}+\cdots\right]$

Let $u = -2x^{2}+\dfrac{2}{3}x^{4}-\dfrac{4}{45}x^{6}+\cdots$. Using $\ln(1+u) = u - \dfrac{u^{2}}{2}+\dfrac{u^{3}}{3}-\cdots$:

$u^{2} = 4x^{4}-\dfrac{8}{3}x^{6}+\cdots$, \;\; $u^{3} = -8x^{6}+\cdots$

$\ln(\cos 2x) = \left(-2x^{2}+\dfrac{2}{3}x^{4}-\dfrac{4}{45}x^{6}\right)-\dfrac{1}{2}\left(4x^{4}-\dfrac{8}{3}x^{6}\right)+\dfrac{(-8x^{6})}{3}+\cdots$

$= -2x^{2}+\left(\dfrac{2}{3}-2\right)x^{4}+\left(-\dfrac{4}{45}+\dfrac{4}{3}-\dfrac{8}{3}\right)x^{6}+\cdots$

$= -2x^{2}-\dfrac{4}{3}x^{4}-\dfrac{64}{45}x^{6}+\cdots$

The expansion is not valid when $\cos 2x \leq 0$. In $0 \leq x \leq \frac{\pi}{4}$: $\cos 2x = 0$ at $x = \dfrac{\pi}{4}$.

$\displaystyle\int\dfrac{\ln(\cos 2x)}{x^{2}}\,\mathrm{d}x \approx \int\dfrac{-2x^{2}-\frac{4}{3}x^{4}-\frac{64}{45}x^{6}}{x^{2}}\,\mathrm{d}x$

$= \displaystyle\int\left(-2-\dfrac{4}{3}x^{2}-\dfrac{64}{45}x^{4}\right)\mathrm{d}x$

$= -2x-\dfrac{4}{9}x^{3}-\dfrac{64}{225}x^{5}+\cdots$

$\displaystyle\int_{0}^{0.5}\dfrac{\ln(\cos 2x)}{x^{2}}\,\mathrm{d}x \approx \left[-2x-\dfrac{4}{9}x^{3}-\dfrac{64}{225}x^{5}\right]_{0}^{0.5}$

$= -\dfrac{479}{450} = -1.0644 \approx -1.064$ (4 d.p.)

Using GC:

$\displaystyle\int_{0}^{0.5}\dfrac{\ln(\cos 2x)}{x^{2}}\,\mathrm{d}x = -1.0670$ (4 d.p.)

N2018 P2 Q4

N2018 P2 Q4

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