# Problem Of The Week #395 Dec 2nd, 2019

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#### anemone

##### MHB POTW Director
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Congratulations to the following members for their correct solution!

1. lfdahl
2. castor28

Solution from lfdahl :
One way to solve the problem is to use the orthogonality of sine functions:

If $n$ and $m$ are positive integers, then

$\int_{0}^{2\pi}\sin mx \sin nx dx = \frac{1}{2}\int_{0}^{2\pi}\left ( \cos (m-n)x - \cos (m+n)x\right )dx =\left\{\begin{matrix} 0, \;\;\;m \ne n\\ \pi, \;\;\; m = n \end{matrix}\right.$

Thus multiplying by $\sin x$ and integrating over the interval $[0;2\pi]$, we get

$\sin^7x = a\sin 7x + b \sin 5x + c \sin 3x + d \sin x$

$\Rightarrow \int_{0}^{2\pi}\sin^8x dx = a\int_{0}^{2\pi}\sin 7x \sin x dx +b\int_{0}^{2\pi}\sin 5x \sin x dx +a\int_{0}^{2\pi}\sin 3x \sin x dx +d\int_{0}^{2\pi}\sin^2x dx \\\\ =d\int_{0}^{2\pi}\sin^2x dx \\\\ = \pi \cdot d$

It remains to validate the definite integral of $\sin^8x$. In doing so, we can use the power reduction formula:

$\int_{0}^{2\pi}\sin^nx dx = \frac{n-1}{n}\int_{0}^{2\pi}\sin^{n-2}xdx - \left [ \frac{1}{n}\cos x \sin^{n-1}x \right ]^{2\pi}_0 = \frac{n-1}{n}\int_{0}^{2\pi}\sin^{n-2}xdx$

- which can be easily shown by integration by parts. Thus, we end up with:

$d= \frac{1}{\pi}\int_{0}^{2\pi}\sin^8x dx = \frac{1}{\pi}\frac{7}{8}\int_{0}^{2\pi}\sin^6x dx = \frac{1}{\pi}\frac{7\cdot 5}{8\cdot6}\int_{0}^{2\pi}\sin^4x dx = \frac{1}{\pi}\frac{7\cdot 5 \cdot 3}{8\cdot6 \cdot 4}\int_{0}^{2\pi}\sin^2x dx \\\\ = \frac{7\cdot 5 \cdot 3}{8\cdot 6 \cdot 4} = \frac{35}{64}.$

Alternate solution from castor28 :
We write the identity for a few values of $x$ (in degrees):
$$\begin{array}{ll} x=30:\qquad&-\dfrac{a}{2}+\dfrac{b}{2}+c+\dfrac{d}{2}=\dfrac{1}{128}\\ x=45: &\dfrac{\sqrt{2}}{2}(-a-b+c+d)=\left(\dfrac{\sqrt{2}}{2}\right)^7=\dfrac{\sqrt{2}}{16}\\ x=60:&\dfrac{\sqrt{3}}{2}(a-b+d)=\left(\dfrac{\sqrt{3}}{2}\right)^7 = \dfrac{27\sqrt{3}}{128}\\ x=90:&-a+b-c+d=1 \end{array}$$
After simplification, we get the system of linear equations:
\begin{align*}
-a+b+2c+d&=\dfrac{1}{64}\\
-a-b+c+d&=\dfrac18\\
a-b+d&=\dfrac{27}{64}\\
-a+b-c+d&=1
\end{align*}

whose solution is $a=-\dfrac{1}{64}$, $b=\dfrac{7}{64}$, $c=-\dfrac{21}{64}$, $\bf d=\dfrac{35}{64}$.

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