Day: October 14, 2022

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202210141629 Solution to 1976-CE-AMATH-II-XX

By using

\displaystyle{\int_{a}^{b}f(x)\,\mathrm{d}x = \int_{a}^{c}f(x)\,\mathrm{d}x + \int_{c}^{b}f(x)\,\mathrm{d}x}

or otherwise, evaluate

\displaystyle{\int_{0}^{\pi}\cos^{7}x\,\mathrm{d}x}.

Warm-up.

\begin{aligned} f(x) & =\cos^7x \\ f(-x) & = \cos^7(-x) \\ & = \cos^7(x) \\ & = f(x) \\ \end{aligned}

\therefore f(x) is an even function such that

\displaystyle{\int_{-a}^{a}f(x)\,\mathrm{d}x = 2\int_{0}^{a}f(x)\,\mathrm{d}x=2\int_{-a}^{0}f(x)\,\mathrm{d}x}.

Assume f(x+T)=f(x) is a periodic function for some T\in (0,2\pi ].

\begin{aligned} & x\overset{f} \longrightarrow \cos^7x \\ \sim\enspace & x\overset{g}\longrightarrow \cos x\overset{h}\longrightarrow \cos^7x \\ \end{aligned}

where g(x)=\cos x, h(x)=x^7, and f=h\circ g.

g:[0,\pi ]\to [-1,1] by x\mapsto \cos x is one-to-one and onto.

h:[-1,1]\to [-1,1] by x\mapsto x^7 is injective and surjective as well.

Hence f, the composition h\circ g of bijections g and h, is also bijective. Besides f:[0,\pi ]\to [-1,1] is a continuously differentiable function that f(0)=1, f(\frac{\pi}{2}) =0, and f(\pi )=-1.

How do you evaluate the integral below:

\displaystyle{F(\theta )=\int_{0}^{\theta}f(x)\,\mathrm{d}x}

where F is an odd function such that

\begin{aligned} F(0) & = F(2\pi ) = 0 \\ F(\theta ) & = -F(-\theta ) \\ \end{aligned}

The curve of function f is flat at point(s) whose slope f' is zero, i.e.,

\begin{aligned} f'(x) & = 0 \\ \frac{\mathrm{d}}{\mathrm{d}x}(\cos^7x) & = 0 \\ -7\cos^6x\sin x & = 0 \\ x & = 0,\frac{\pi}{2}, \pi \\ \end{aligned}

Roughwork. Show

    Integrating by parts,

    \begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}\cos^6x\,\mathrm{d}(\sin x) \\ & = \big[\cos^6x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^6x)\\ & = 0 + \int_{0}^{\pi}6\sin^2x\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6(1-\cos^2x)\cos^5x\,\mathrm{d}x \\ & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x - \int_{0}^{\pi}6\cos^7x\,\mathrm{d}x \\ \int_{0}^{\pi}7\cos^7x\,\mathrm{d}x & = \int_{0}^{\pi}6\cos^5x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x \\ & = \frac{6}{7}\int_{0}^{\pi}\cos^4x\,\mathrm{d}(\sin x) \\ &= \frac{6}{7}\bigg\{ \big[\cos^4x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^4x)\bigg\} \\ &= \frac{6}{7}\bigg\{ 0 + \int_{0}^{\pi}4\sin^2 x\cos^3x\,\mathrm{d}x\bigg\} \\ & = \frac{24}{7}\int_{0}^{\pi} (1-\cos^2x)\cos^3x\,\mathrm{d}x \\ \frac{30}{7}\int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{24}{7} \int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^5x\,\mathrm{d}x & = \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \\ \end{aligned}

    \begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{6}{7}\bigg\{ \frac{4}{5}\int_{0}^{\pi}\cos^3x\,\mathrm{d}x \bigg\} \\ & = \frac{24}{35} \int_{0}^{\pi} \cos^2x\,\mathrm{d}(\sin x) \\ & = \frac{24}{35}\bigg\{ \big[ \cos^2x\sin x\big]\big|_{0}^{\pi} - \int_{0}^{\pi}\sin x\,\mathrm{d}(\cos^2x) \bigg\} \\ & = \frac{24}{35} \bigg\{ 0 + \int_{0}^{\pi}2\sin^2x\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{48}{35} \int_{0}^{\pi}(1-\cos^2x)\cos x\,\mathrm{d}x \\ \bigg(\frac{24}{35}+\frac{48}{35}\bigg) \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{48}{35}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi} \cos^3x\,\mathrm{d}x & = \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \\ \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \frac{24}{35}\bigg\{ \frac{2}{3}\int_{0}^{\pi}\cos x\,\mathrm{d}x \bigg\} \\ & = \frac{16}{35}\big[ \sin x\big]\big|_{0}^{\pi} \\ & = 0 \\ \end{aligned}

    Solution.

    \begin{aligned} \int_{0}^{\pi}\cos^7x\,\mathrm{d}x & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{\frac{\pi}{2}}^{\pi}\cos^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x + \int_{0}^{\frac{\pi}{2}}\cos^7(x-\pi /2)\,\mathrm{d}(x-\pi /2) \\ & = \int_{0}^{\frac{\pi}{2}}\cos^7x\,\mathrm{d}x - \int_{0}^{\frac{\pi}{2}}\sin^7x\,\mathrm{d}x \\ & = \int_{0}^{\frac{\pi}{2}}(\cos^7x-\sin^7x)\,\mathrm{d}x \\ \end{aligned}

    The rest is left as an exercise to the reader.

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202210141016 Solution to 1975-CE-AMATH-II-XX

In the Figure below, OABC is a square. D and E are the mid-points of AB and BC respectively. Let the vector \overrightarrow{OA} be represented by \mathbf{i} and \overrightarrow{OC} by \mathbf{j}.

(a) Express \overrightarrow{OD} in terms of \mathbf{i} and \mathbf{j}.
(b) Express \overrightarrow{OE} in terms of \mathbf{i} and \mathbf{j}.
(c) Evaluate \displaystyle{\frac{\overrightarrow{OD}\cdot\overrightarrow{OE}}{\left|\overrightarrow{OD}\right|\left|\overrightarrow{OE}\right|}}, where \left|\overrightarrow{OD}\right| and \left|\overrightarrow{OE}\right| are the magnitudes of \overrightarrow{OD} and \overrightarrow{OE} respectively.
(d) Hence calculate \angle DOE.

Notation.

In what follow vectors will be typographically boldfaced, i.e.,

\begin{aligned} \mathbf{OA} & := \overrightarrow{OA} \\ \mathbf{OC} & := \overrightarrow{OC} \\ \mathbf{OD} & := \overrightarrow{OD} \\ \mathbf{OE} & := \overrightarrow{OE} \\ \end{aligned}

etc., in place of an overhead arrow \overrightarrow{[\cdot ]} more than usually handwritten, the direction of which (e.g. \overrightarrow{AB}) is intended an initial point (e.g. A) make to a terminal point (e.g. B).

(a)

\begin{aligned} \mathbf{OD} & = \mathbf{OA} + \mathbf{AD} \\ & = \mathbf{OA} + \frac{1}{2}\mathbf{AB} \\ \dots\enspace\because\enspace & \mathbf{AB}=\mathbf{OC}\enspace\dots \\ & = \mathbf{OA} + \frac{1}{2}\mathbf{OC} \\ & = \mathbf{i} + \frac{1}{2}\mathbf{j} \\ \end{aligned}

(b)

\begin{aligned} \mathbf{OE} & = \mathbf{OC} + \mathbf{CE} \\ & = \mathbf{OC} + \frac{1}{2}\mathbf{CB} \\ \dots\enspace\because\enspace & \mathbf{CB}=\mathbf{OA}\enspace\dots \\ & = \mathbf{OC} + \frac{1}{2}\mathbf{OA} \\ & = \mathbf{j} + \frac{1}{2}\mathbf{i} \\ \end{aligned}

(c) and (d) are not chosen.