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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.