0-pi-2-x-sin-8-x-cos-8-x-dx-

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

\int_{0}^{\frac{π}{2}} \frac{x}{\sin^{8} x + \cos^{8} x} d x ?

Answered by phanphuoc last updated on 23/Oct/21

\int_{0}^{π / 2} \frac{π / 2 - x}{{s i n}^{8} (π / 2 - x) + {c o s}^{8} (π / 2 - x)} d x =

= \int_{0}^{π / 2} \frac{π / 2 - x}{{s i n}^{8} x + {c o s}^{8} x} d x

\to i = π / 4 \int_{0}^{π / 2} \frac{d (t a n x)}{{t a n}^{8} x + 1}

= π / 4 \int_{0}^{\infty} \frac{d t}{t^{8} + 1} = π / 4 . (π / 8 s i n (π / 8) = π^{2} / 32 s i n (π / 8)

Commented by cortano last updated on 23/Oct/21

t h e r e s u l t i s \frac{π^{2}}{8 \sqrt{2}} (\sqrt{2 + \sqrt{2}} + 3 \sqrt{2 - \sqrt{2}})

Answered by phanphuoc last updated on 23/Oct/21

u p d a t e

- \int_{0}^{\infty} d x / (x^{n} + 1) = \frac{π}{n} s i n \frac{π}{n}

y o u c a n w a t c h s y p u t u b e