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Question Number 157412 by cortano last updated on 23/Oct/21

$${\int }_0^{\frac{\pi }{2}}\frac{x}{\sin {}^{8}x+\cos {}^{8}x}dx?$$

Answered by phanphuoc last updated on 23/Oct/21

$${\int }_0^{\pi /2}\frac{\pi /2-x}{{sin}^8(\pi /2-x)+{cos}^8(\pi /2-x)}dx=$$$$={\int }_0^{\pi /2}\frac{\pi /2-x}{{sin}^{8}x+{cos}^{8}x}dx$$$$\to i=\pi /4{\int }_0^{\pi /2}\frac{d\left(tanx\right)}{{tan}^{8}x+1}$$$$=\pi /4{\int }_0^{\mathrm{\infty }}\frac{dt}{t^8+1}=\pi /4.(\pi /8sin\left(\pi /8\right)={\pi }^2/32sin\left(\pi /8\right)$$

Commented by cortano last updated on 23/Oct/21

$$theresultis\frac{{\pi }^2}{8\sqrt{2}}(\sqrt{2+\sqrt{2}}+3\sqrt{2-\sqrt{2}})$$

Answered by phanphuoc last updated on 23/Oct/21

$$update$$$$-{\int }_0^{\mathrm{\infty }}dx/(x^n+1)=\frac{\pi }{n}sin\frac{\pi }{n}$$$$youcanwatchsyputube$$

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