Question Number 153104 by amin96 last updated on 04/Sep/21
$$\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} +\left(\pi−{x}\right)^{\mathrm{3}} }{dx}=? \\ $$
Answered by puissant last updated on 04/Sep/21
![I=∫_0 ^π (x^3 /(x^3 +(π−x)^3 ))dx (1) u=π−x → dx=−du. I=∫_0 ^π (((π−u)^3 )/((π−u)^3 +u^3 ))du (2) (((1)+(2))/2) ⇒ I=(1/2)[x]_0 ^π ∴∵ I=(π/2)..](Q153122.png)
$${I}=\int_{\mathrm{0}} ^{\pi} \frac{{x}^{\mathrm{3}} }{{x}^{\mathrm{3}} +\left(\pi−{x}\right)^{\mathrm{3}} }{dx}\:\left(\mathrm{1}\right) \\ $$$${u}=\pi−{x}\:\rightarrow\:{dx}=−{du}. \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} \frac{\left(\pi−{u}\right)^{\mathrm{3}} }{\left(\pi−{u}\right)^{\mathrm{3}} +{u}^{\mathrm{3}} }{du}\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)+\left(\mathrm{2}\right)}{\mathrm{2}}\:\Rightarrow\:{I}=\frac{\mathrm{1}}{\mathrm{2}}\left[{x}\right]_{\mathrm{0}} ^{\pi} \\ $$$$ \\ $$$$\therefore\because\:\:\:{I}=\frac{\pi}{\mathrm{2}}.. \\ $$