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Please point me out some useful notes for evaluating the following integral. I really wonder why it is $0$. $\int_{-0.5}^{0.5}\cos(x)\ln\frac{1+x}{1-x}dx=0$ I apply the rules which I know about solving the integral but, they have been useless. Any help will be appreciated.

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    I was drawing the function when the following answer came. Thanks @Sasha.2012-08-02

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The reason is this: if $f$ is an odd function then $\displaystyle \int_{-L}^L f(x)\, dx = 0$ for any real number $L$.

Explicitly in this case: we can split the integral as

$\int_0^{\frac{1}{2}} \cos x \ln \dfrac{1+x}{1-x}\, dx + \int_{-\frac{1}{2}}^0 \cos x \ln \dfrac{1+x}{1-x}\, dx$

Substituting $u=-x$ into the second integral then gives

$\int_0^{\frac{1}{2}} \cos x \ln \dfrac{1+x}{1-x}\, dx + \int_{\frac{1}{2}}^0 \cos (-u) \ln \dfrac{1-u}{1+u}\, (-du)$

Then using symmetry properties, namely $\cos(-u)=\cos u$ and $\ln \dfrac{1-u}{1+u} = -\ln \dfrac{1+u}{1-u}$, we see that the second integral is the negative of the first, so they cancel to give zero.

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    If it wasn't very apparent: the logarithmic factor is in fact twice the inverse of the hyperbolic tangent, which is known to be odd. We are then integrating the product of an odd function and an even function over a symmetric interval...2012-08-03