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Find this value $$I_{n}=\int_{-1}^{1}\arccos{\left(\sum_{k=1}^{n}(-1)^{k-1}x^{2k-1}\right)}dx=\pi?$$

My try: since $$\sum_{k=1}^{n}(-1)^{k-1}x^{2k-1}=\dfrac{x(1-(-x^2)^{n})}{1+x^2}$$ so $$I_{n}=\int_{-1}^{1}\arccos{\left(\dfrac{x(1-(-x^2)^{n})}{1+x^2}\right)}dx$$ then I can't work.Thank you very much !

math110
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  • Have you tried stupidly integrating by parts? What shows up? When $n=1$ integration by parts gives you $\pi$ as the answer in one step. (I don't have paper, I can't try it myself.) – Patrick Da Silva Nov 27 '13 at 17:16
  • Better to write your term as $(-1)^nx^{2n}$. You could start with Patrick's suggestion and try induction; you might want to split that into two cases -- n even, n odd. – Betty Mock Nov 27 '13 at 18:25

1 Answers1

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Let $\displaystyle\phi(x)=\cos^{-1}\left(\sum_{k=0}^{n-1}(-1)^kx^{2k+1}\right)$. Notice that $\phi(-x)+\phi(x)=\pi$.

We are looking for $$ \begin{align} \int_{-1}^1\phi(x)\,\mathrm{d}x &=\int_0^1(\phi(x)+\phi(-x))\,\mathrm{d}x\\ &=\int_0^1\pi\,\mathrm{d}x\\[9pt] &=\pi \end{align} $$ We could have used any odd function that maps $[-1,1]\mapsto[-1,1]$ in place of the sum we used here and get the same answer.

robjohn
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