Prove that $\displaystyle\int_0^1 \dfrac{x^{2n}}{x^{2n}+{(1-x)^{2n}}}dx=\dfrac{1}{2}$ I have only proved that $\displaystyle\int_0^1 \dfrac{x^{2}}{x^{2}+{(1-x)^{2}}}dx=\dfrac{1}{2}$
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replace $x$ with $x+\frac{1}{2}$ and the integral becomes $\int_{-0.5}^{0.5}\frac{1}{2}dx$. – Callus - Reinstate Monica Mar 09 '17 at 19:40
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Making the substitution $x\to1-x$, we get $$ \int_0^1\frac{x^{2n}}{x^{2n}+(1-x)^{2n}}\;dx=\int_0^1\frac{(1-x)^{2n}}{x^{2n}+(1-x)^{2n}}\;dx$$ and since $$ \int_0^1\frac{x^{2n}}{x^{2n}+(1-x)^{2n}}\;dx+\int_0^1\frac{(1-x)^{2n}}{x^{2n}+(1-x)^{2n}}\;dx=\int_0^1\;dx=1$$ it follows that $$ \int_0^1\frac{x^{2n}}{x^{2n}+(1-x)^{2n}}\;dx=\frac{1}{2} $$
carmichael561
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Generally Suppose we want to find
$$I(a) = \int^a_0 \frac{f(x)}{f(x)+f(a-x)} \mathrm{d}x $$
Then since $$\int^a_0 \frac{f(x)}{f(x)+f(a-x)} \mathrm{d}x = \int^a_0 \frac{f(a-x)}{f(x)+f(a-x)}\mathrm{d}x$$
$$2I(a) = \int^a_0 \frac{f(x)+f(a-x)}{f(x)+f(a-x)} \mathrm{d}x = \int^a_0 \mathrm{d}x = a$$
Hence
$$I(a) = \int^a_0 \frac{f(x)}{f(x)+f(a-x)} \mathrm{d}x = \frac{a}{2}$$
Zaid Alyafeai
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