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Show that

$$\int_{0}^{\pi\over 2}(\sin{x}-\cos{x})\ln{\left({\sin{x}\over \sin{x}+\cos{x}}\right)}\mathrm dx=\color{blue}{\ln{2}}.\tag1$$

Enforcing $u=\sin{x}+\cos{x}$ the $du=\cos{x}-\sin{x}dx$

$$\int_{1}^{1}\ln{\left({\sin{x}\over u}\right)}\mathrm du=\color{blue}{\ln{2}}.$$

The variable x is still there. I can't seem to think of away to get rid of it. I needn't help.

Can anyone help to prove $(1)?$

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    How can the integral "from $1$ to $1$" give a nonzero result? – 2017-01-12
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    It is getting boring: you just need to exploit symmetry. Again. $x\to\frac{\pi}{2}-x$. – 2017-01-12
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    The purpose of MSE is to share knowledge and help people. If you keep asking questions that can be solved by using the same trick over and over, and you seem to not learn anything from our answers, you are kind of abusing our time, don't you think? We are not a "solve this for free" service. – 2017-01-12
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    @JackD'Aurizio Well said. Also I advise you @ abdi to moderate your words. As Jack said, we are not a "solve for free" service, and your insults wouldn't make us to tolerate you more than how much already we do not. – 2017-01-12
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    if you are not able to follow the symmetry arguments given frequently to answer your (exhaustingly repetitve) questions you may try integration by parts which works quite well for integrals like this – 2017-01-12

1 Answers 1

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Let $$I=\int_{0}^{\pi\over 2}(\sin{x}-\cos{x})\ln{\left({\sin{x}\over \sin{x}+\cos{x}}\right)}\,\mathrm dx$$ and let $x\to\dfrac{\pi}{2}-x$ we get $$I=-\int_{0}^{\pi\over 2}(\sin{x}-\cos{x})\ln{\left({\cos{x}\over \sin{x}+\cos{x}}\right)}\,\mathrm dx$$ hence $$2I=\int_{0}^{\pi\over 2}(\sin{x}-\cos{x})\ln{\tan x}\,\mathrm dx=\int_{0}^{\pi\over 2}\sin{x}\ln{\tan x}\,\mathrm dx-\int_{0}^{\pi\over 2}\cos{x}\ln{\tan x}\,\mathrm dx$$ these two integrals are not hard to calculate, hope you can take it from here.