$$\int_{0}^{\frac{\pi}{2}} \frac {1-\cos(2nx) }{1-\cos(2x) } dx = n\frac {\pi}{2} $$
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Related :http://math.stackexchange.com/questions/263705/compute-int-0-pi-2-frac-sin-2013x-sin-x-dx-space – lab bhattacharjee Mar 10 '17 at 11:53
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Also : http://math.stackexchange.com/questions/1034156/ways-to-prove-displaystyle-int-0-pi-dx-dfrac-sin2n-x-sin2-x-n-pi/1034663#1034663 – lab bhattacharjee Mar 10 '17 at 11:58
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Consider \begin{align*} \int_{0}^{\pi}\frac{1-e^{\iota nx}}{1-e^{\iota x}}\mathrm{d}{x} &=\int_{0}^{\pi}\left(1+\cdots+e^{\iota(n-1)x}\right)\mathrm{d}{x} \\&=\pi+\iota\sum\limits_{k=1}^{n-1}\frac{1-e^{\iota k\pi}}{k} \\&=\pi+\iota\sum\limits_{k=1}^{n-1}\frac{1-(-1)^{k}}{k} \end{align*} Then \begin{align*} 2\Re\int_{0}^{\pi}\frac{1-e^{\iota nx}}{1-e^{\iota x}}\mathrm{d}{x} &=\int_{0}^{\pi}\left(\frac{1-e^{\iota nx}}{1-e^{\iota x}}+\frac{1-e^{-\iota nx}}{1-e^{-\iota x}}\right)\mathrm{d}{x} \\&=\int_{0}^{\pi}\frac{1-\cos{(nx)}-\cos{(x)}+\cos{((n-1)x)}}{1-\cos{(x)}}\mathrm{d}{x} \end{align*} Hence, \begin{align*} \int_{0}^{\pi}\frac{1-\cos{\left(nx\right)}}{1-\cos{\left(x\right)}}\mathrm{d}{x} =\pi+\int_{0}^{\pi}\frac{1-\cos{\left((n-1)x\right)}}{1-\cos{\left(x\right)}}\mathrm{d}{x} =\cdots=n\pi \end{align*}
shdp
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I think a similar reasoning applies, give it a try, I'll write it up in a minute – shdp Mar 10 '17 at 12:09
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Im a 12th grade student , a guy gave me that question and he asked me to prove it . I actually don't understand what math you've done there . – DocDev Mar 10 '17 at 14:42
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There is no question there, just a definite integral, that is not an equation. A proof would be in order if there was an statement there, such as $$\int_{0}^{\frac{\pi}{2}}\frac{1-\cos{\left(nx\right)}}{1-\cos{\left(x\right)}}\mathrm{d}{x}=\mbox{ Something}$$ – shdp Mar 10 '17 at 17:12