# Helpful Hints to solve a difficult Integral

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.

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!

• Hint: If you plot the integrand, you get something asymmetric on $[-\pi/2,\pi/2]$. Try applying symmetry. Aug 22, 2019 at 16:02
• 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: $$2I=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}x^2 \cos xdx$$ Aug 22, 2019 at 16:04

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.
\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 $$14.$$