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

$${\int }_0^{\pi }\frac{x^3}{x^3+{(\pi -x)}^3}dx=?$$

Answered by puissant last updated on 04/Sep/21

$$I={\int }_0^{\pi }\frac{x^3}{x^3+{(\pi -x)}^3}dx\left(1\right)$$$$u=\pi -x\to dx=-du.$$$$I={\int }_0^{\pi }\frac{{(\pi -u)}^3}{{(\pi -u)}^3+u^3}du\left(2\right)$$$$\frac{\left(1\right)+\left(2\right)}{2}\Rightarrow I=\frac{1}{2}{\left[x\right]}_0^{\pi }$$$$$$$$\therefore \because I=\frac{\pi }{2}..$$

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