# Evaluating an Integral makes it zero

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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Plot the function over the domain. Notice the (anti-)symmetry, draw conclusion. –  Sasha Aug 2 '12 at 22:01
What happens when you replace $x$ with $-x$ in $\cos(x)\ln\frac{1+x}{1-x}$? –  Antonio Vargas Aug 2 '12 at 22:04
Another hint: $\log y = -\log 1/y$. –  Antonio Vargas Aug 2 '12 at 22:05
I was drawing the function when the following answer came. Thanks @Sasha. –  Ned Dabby Aug 2 '12 at 22:07

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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I got the point. Thanks Clive. –  Ned Dabby Aug 2 '12 at 22:09
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... –  Ｊ. Ｍ. Aug 3 '12 at 3:41