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$$ \int_{-\pi/4}^{\pi/4}\frac{\cos x}{1+e^x}dx $$ I tried to calculate this integration by substitution $x$ by $-x$ and the two integration

Arthur
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Mario
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  • Use https://math.stackexchange.com/questions/439851/evaluate-the-integral-int-frac-pi2-0-frac-sin3x-sin3x-cos3x/439856#439856 – lab bhattacharjee Nov 27 '19 at 07:21
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    What you've described is exactly the way to solve this problem. Can you show us where you got stuck? – JimmyK4542 Nov 27 '19 at 07:21
  • I substituted x by -x and i added the new integral with the first integral – Mario Nov 27 '19 at 07:42
  • Then i divided the whole integral by 2 and i calculated it and i substituded by the boundry – Mario Nov 27 '19 at 07:42
  • Wolfram alpha gives $$\frac{1}{2} \sqrt[4]{-1} \left(-i , _2F_1\left(-i,1;1-i;-e^{-\pi /4}\right)+, _2F_1\left(-i,1;1-i;-e^{\pi /4}\right)+, _2F_1\left(i,1;1+i;-e^{-\pi /4}\right)-i , _2F_1\left(i,1;1+i;-e^{\pi /4}\right)\right)$$ – Dr. Sonnhard Graubner Nov 27 '19 at 07:45

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You had the good intuition : $$ 2I = \int_{-\frac{\pi}{4}}^{+\frac{\pi}{4}} \frac{\cos x}{1+\mathrm{e}^x}\mathrm{d}x + \int_{-\frac{\pi}{4}}^{+\frac{\pi}{4}} \frac{\cos -x}{1+\mathrm{e}^{-x}}\mathrm{d}x = \int_{-\frac{\pi}{4}}^{+\frac{\pi}{4}} \frac{(2+\mathrm{e}^{-x}+\mathrm{e}^{x})\cos x}{(1+\mathrm{e}^{-x})(1+\mathrm{e}^{x})}\mathrm{d}x $$ So,

$$ I = \frac{1}{2}\int_{-\frac{\pi}{4}}^{+\frac{\pi}{4}}\cos x\mathrm{d}x = \frac{\sqrt{2}}{2} $$

Mister Da
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