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I am attempting to solve this integral/problem I found on brilliant. Given$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{x^{2}\cos(x)}{1+\exp(x^{2}\sin(x))}\,dx $$ converges to $\dfrac{\pi^{a}-b}{c}$, with $a,b,c \in \mathbb{Z}$, find $a+b+c.$ I was overly curious and spoiled the surprise by going to wolframalpha, but I would like some hints to solve the problem analytically without this helpful computation engine.

For context: https://www.wolframalpha.com/input/?i=integrate&assumption=%7B%22C%22%2C+%22integrate%22%7D+-%3E+%7B%22Calculator%22%7D&assumption=%7B%22F%22%2C+%22Integral%22%2C+%22rangestart%22%7D+-%3E%22-pi%2F2%22&assumption=%7B%22F%22%2C+%22Integral%22%2C+%22integrand%22%7D+-%3E%22%5B%5Bx%5E2%5Dcos%28x%29%5D%2F%5B1%2Be%5E%28%28x%5E2%29sin%28x%29%29%5D%22&assumption=%7B%22F%22%2C+%22Integral%22%2C+%22rangeend%22%7D+-%3E%22pi%2F2%22

https://www.wolframalpha.com/input/?i=0.467401&assumption=%22ClashPrefs%22+-%3E+%7B%22Math%22%7D

Link $1$ is the evaluated integral, Link $2$ gives the answer in terms of $\pi$ and we let $a=2, b=8, c=4$ and the sum is $14.$ All hints to get initially started towards a solution are greatly appreciated. I tried a $u$ substitution and it was absolutely bologna. Feynman integration or by parts perhaps? Please help!

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  • $\begingroup$ Hint: If you plot the integrand, you get something asymmetric on $[-\pi/2,\pi/2]$. Try applying symmetry. $\endgroup$ Aug 22, 2019 at 16:02
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    $\begingroup$ Hint: Let $x\to -x$: $$I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{x^{2}\cos x}{1+\exp(x^{2}\sin x )}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{x^{2}\cos x}{1+\exp(-x^{2}\sin x)}dx$$ Adding the two integrals from above gives: $$2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^2 \cos xdx$$ $\endgroup$
    – Zacky
    Aug 22, 2019 at 16:04

2 Answers 2

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Note that this is of the form $$\int_{-a}^a\frac{\text{even(x)}}{1+b^{\text{odd}(x)}}\mathrm{d}x=\int_0^a \text{even}(x)\mathrm{d}x$$ for arbitrary even/odd functions and constants $a,b\in\mathbb{R}^+$. A proof of the above can be found here.

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I ended up using Peter Foreman's suggestion. The derivation was quite straight-forward:

\begin{align*} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\dfrac{x^{2}\cos(x)}{1+\exp(x^{2}\sin(x))}\,dx &= \int_{0}^{\frac{\pi}{2}}x^{2}\cos(x)\,dx \\ &=x^{2}\sin(x)\,\bigg\rvert_{0}^{\frac{\pi}{2}}-\int_{0}^{\frac{\pi}{2}}2x\,\sin(x)\,dx \\ &=\dfrac{\pi^{2}}{4}-\int_{0}^{\frac{\pi}{2}}2x\,\sin(x)\,dx \\ &=\dfrac{\pi^{2}}{4}-\bigg[-2x\cos(x)\bigg\rvert_{0}^{\frac{\pi}{2}}+\int_{0}^{\frac{\pi}{2}}2\cos(x)\,dx\bigg]\\ &=\dfrac{\pi^{2}}{4}-\bigg[0+2\sin(x)\bigg\rvert_{0}^{\frac{\pi}{2}}\bigg]\\ &=\dfrac{\pi^{2}}{4}-2 \\ &=\dfrac{\pi^{2}-8}{4}. \end{align*} Thus we have $a=2, b=8, c=4$ and the sum of these integers is $14.$

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