Question

The value of the integral
$\frac{48}{\pi^4} \int_{0}^{\pi}(\frac{3\pi x^2}{2} - x^3) \frac{ \sin(x)}{1 + \cos^2x} \, dx$
is equal to ________

Answer 1

The correct answer is 6
$I = \frac{48}{\pi^4} \int_{0}^{\pi}\left[\frac{\pi}{2} - x\right)^3 - \frac{3\pi^2}{4}(\frac{\pi}{2} - x) + \frac{\pi^3}{4}] \frac{ \sin(x)dx}{1 + \cos^2(x)}$
Using
$\int_{a}^{b} f(x) \, dx = \int_{a}^{b} f(a + b - x) \, dx$
we get
$I = \frac{48}{\pi^4} \int_{0}^{\pi}\left[\frac{\pi}{2} - x\right)^3 + \frac{3\pi^2}{4}(\frac{\pi}{2} - x) + \frac{\pi^3}{4}] \frac{ \sin(x)dx}{1 + \cos^2(x)}$
Adding these two equations, we get
$2I = \frac{48}{\pi^4} \int_{0}^{\pi}\frac{\pi^{3}}{2}. \frac{ \sin(x)dx}{1 + \cos^2(x)}$
$⇒I=\frac{12}{π}[−tan^{−1}⁡(cos⁡x)]^{π}_{0}$
$=\frac{12}{π}⋅\frac{π}{2}=6$