0-pi-x-3-x-3-pi-x-3-dx-

Question Number 153104 by amin96 last updated on 04/Sep/21

\int_{0}^{π} \frac{x^{3}}{x^{3} + {(π - x)}^{3}} d x = ?

Answered by puissant last updated on 04/Sep/21

I = \int_{0}^{π} \frac{x^{3}}{x^{3} + {(π - x)}^{3}} d x (1)

u = π - x \to d x = - d u .

I = \int_{0}^{π} \frac{{(π - u)}^{3}}{{(π - u)}^{3} + u^{3}} d u (2)

\frac{(1) + (2)}{2} \Rightarrow I = \frac{1}{2} {[x]}_{0}^{π}

∴∵ I = \frac{π}{2} . .

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