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The following integrals are the Brasilian Mathematical Olympiad in 2019.

Show that $\displaystyle\int_1^2\frac{e^x(x-1)}{x(x+e^x)}\,dx=\ln\left(\frac{2+e^2}{2+2e}\right)$ and $\displaystyle\int_0^{\pi/2}\frac{x\sin(2x)}{1+\cos(2x)^2}\,dx=\frac{\pi^2}{16}$.

I tried to use traditional methods, but these integrals are very difficult. Has someone any suggestion? Thanks a lot.

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    $\begingroup$ As a "test-taker's answer" for the first one, the fact that $F(2) - F(1) = \ln \left(\frac{2 + e^2}{2 + 2e}\right)$ is a big hint that $F(x)$ is probably $\ln \left(\frac{x + e^x}{x + xe}\right)$ (such that $F(2) = \ln \left(\frac{2 + e^2}{2 + 2e}\right)$ and $F(1) = 0$); and it's straightforward to confirm that that is in fact the case, by taking its derivative, $F^\prime(x) = \frac{e^x(x-1)}{x(x+e^x)}$. $\endgroup$ – ruakh Sep 28 '19 at 1:38
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$$\int_1^2\frac{e^x(x-1)}{x(x+e^x)}dx=\int_1^2 \frac{e^x\left(\frac{1}{x}-\frac{1}{x^2}\right)}{1+\frac{e^x}{x}}dx=\ln\left(1+\frac{e^x}{x}\right)\bigg|_1^2 =\ln\left(\frac{1+\frac{e^2}{2}}{1+e}\right)$$

For the second one, let $\frac{\pi}{2}-x=t$ to get: $$\mathcal J=\int_0^{\pi/2}\frac{x\sin(2x)}{1+\cos^2(2x)}dx=\int_{\pi/2}^0 \frac{\left(\frac{\pi}{2}-t\right)\sin(2t)}{1+\cos^2(2t)}(-dt)\overset{t=x}=\int_0^\frac{\pi}{2}\frac{\left(\frac{\pi}{2}-x\right)\sin(2x)}{1+\cos^2(2x)}dx$$ Now add them up to get: $$2\mathcal J=\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{\sin(2x)}{1+\cos^2(2x)}dx\Rightarrow \mathcal J=-\frac{\pi}{8}\arctan(\cos(2x))\bigg|_0^\frac{\pi}{2}=\frac{\pi^2}{16}$$

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  • $\begingroup$ Your last integral has the wrong bounds of integration after substitution. Just take out the negative. $\endgroup$ – Andrew Chin Sep 24 '19 at 14:42
  • $\begingroup$ If $x\to0$ in $\frac{\pi}2-x=t$, then what is $t\to$? $\endgroup$ – Andrew Chin Sep 24 '19 at 14:43
  • $\begingroup$ @AndrewChin Are you overlooking $dt=-dx$? $\endgroup$ – saulspatz Sep 24 '19 at 14:45
  • $\begingroup$ Oh that's what I overlooked $\endgroup$ – Andrew Chin Sep 24 '19 at 14:46
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For the second integral, use $\int_a^bf(x) = \int_a^bf(a+b-x)$ and add these together to get $I =$$ \int_0^\frac\pi2 {\frac \pi4\sin(2x)\over1+\cos(2x)^2}$. Substituting with $\cos(2x) = t$ gives the solution easily.

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    $\begingroup$ this answer would be a lot more impressive if it was written by a dog $\endgroup$ – clathratus Sep 26 '19 at 22:03
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Hint: $$\int \frac {2\sin (2x)}{1+\cos^2(2x)}dx=-\arctan (\cos(2x))$$ (you can show this by substitution)

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