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Let $\displaystyle I_{n}\!=\!\int^{\pi}_0\frac{4-\cos(n\!-\!1)x-2\cos(nx)-\cos(n\!+\!1)x}{1-\cos x}\,\mathrm dx\\[4pt]$ Then prove that $\displaystyle I_{n+2}+I_n=2I_{n+1}$

My Try :

Using $\displaystyle\;\cos C+\cos D =2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$

$\displaystyle I_{n}=\int^\pi_0\frac{4-[\cos(n+1)x+\cos(n-1)x]-2\cos(nx)}{1-\cos x}\,\mathrm dx=\\[10pt]$ $\displaystyle=\int^\pi_0\frac{4-2\cos(nx)\cos(x)-2\cos(nx)}{1-\cos x}\,\mathrm dx\,.$

$\displaystyle I_n=\int^\pi_0\frac{4-2\cos(nx)(1+\cos x)}{1-\cos x}\,\mathrm dx=\\[10pt]$ $\displaystyle=\int^\pi_0\frac{4-4\cos^2(\frac{x}{2})\cos(nx)}{2\sin^2\left(\frac{x}{2}\right)}\,\mathrm dx=\\[10pt]$ $\displaystyle=\int^\pi_0\frac{2-2\cos^2\left(\frac{x}{2}\right)\cos(nx)}{\sin^2\left(\frac{x}{2}\right)}\,\mathrm dx$

I did not understand how I can reduce that fraction

Please look for that problem. Thanks.

Angelo
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juantheron
  • 53,015

2 Answers2

4

As a hint: $$\displaystyle I_{n}\!=\!\int^{\pi}_0\frac{4-\cos(n\!-\!1)x-2\cos(nx)-\cos(n\!+\!1)x}{1-\cos x}\,\mathrm dx\\[4pt] \\\displaystyle I_{n+2}\!=\!\int^{\pi}_0\frac{4-\cos(n\!+\!1)x-2\cos((n+2)x)-\cos(n\!+\!3)x}{1-\cos x}\,\mathrm dx\\[4pt] $$in $I_{n+2}+I_{n}$ denominator is $1-\cos x$ so go for the sum of two fractions in integrals and simplify like what you have tried before $$\displaystyle=\int^\pi_0\frac{2-2\cos^2(\frac{x}{2})\cos(nx)+2-2\cos^2(\frac{x}{2})\cos((n+2)x)}{\sin^2(\frac{x}{2})}\,\mathrm dx$$ factor $2\cos^2(\frac{x}{2})$ then apply $\displaystyle\;\cos C+\cos D =2\cos\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right)$ for the rest, then simplify and factor $2$.

ACB
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Khosrotash
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3

By using $\;\;\color{brown}{\cos C+\cos D =2\cos\left(\dfrac{C+D}2\right)\cos\left(\dfrac{C-D}2\right)}\,,$

we get that

\begin{align}I_{n+2}+I_n&=\int^\pi_0\frac{4-\cos(n\!+\!1)x-2\cos(n\!+\!2)x-\cos(n\!+\!3)x}{1-\cos x}\,\mathrm dx\,+\\[6pt] &+\int^\pi_0\frac{4-\cos(n\!-\!1)x-2\cos(nx)-\cos(n\!+\!1)x}{1-\cos x}\,\mathrm dx=\\[6px] &=2\!\int^\pi_0\frac{4\!-\!\cos x\big[\!\cos(nx)\!+\!2\cos(n\!+\!1)x\!+\!\cos(n\!+\!2)x\big]}{1-\cos x}\,\mathrm dx=\\[6pt]&=2\!\int^\pi_0\frac{4-\cos(nx)-2\cos(n\!+\!1)x-\cos(n\!+\!2)x}{1-\cos x}\;\mathrm dx\,+\\[6pt]&+2\!\int^\pi_0\!\frac{(1\!-\!\cos x)\big[\cos(nx)\!+\!2\cos(n\!+\!1)x\!+\!\cos(n\!+\!2)x\big]}{1-\cos x}\,\mathrm dx=\\[6pt]&=2I_{n+1}+2\!\int^\pi_0\!\big[\!\cos(nx)\!+\!2\cos(n\!+\!1)x\!+\!\cos(n\!+\!2)x\big]\mathrm dx=\\[6pt]&=2I_{n+1}+2\left[\dfrac{\sin(nx)}n+2\dfrac{\sin(n+1)x}{n+1}+\dfrac{\sin(n+2)x}{n+2}\right]^\pi_0=\\[6pt]&=2I_{n+1}\,.\end{align}

Angelo
  • 12,328