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

Question Number 79763 by mathmax by abdo last updated on 28/Jan/20

c a l c u l a t e \int_{0}^{π} {{c o s}^{8} x + {s i n}^{8} x} d x

Commented by john santu last updated on 28/Jan/20

\cos^{8} x + \sin^{8} x = {(\sin^{4} x)}^{2} + {(\cos^{4} x)}^{2}

= {(\sin^{4} x + \cos^{4} x)}^{2} - 2 \sin^{4} xcos^{4} x

= {{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} xcos^{2} x}^{2} - \frac{1}{2} \sin^{2} (2 x)

= {1 - \frac{1}{2} \sin^{2} (2 x)}^{2} - \frac{1}{2} \sin^{2} (2 x)

= 1 - \frac{3}{2} \sin^{2} (2 x) + \frac{1}{4} \sin^{4} (2 x)

= 1 - \frac{3}{2} (\frac{1}{2} - \frac{1}{2} \cos (4 x)) + \frac{1}{4} {\frac{1}{2} - \frac{1}{2} \cos (4 x)}^{2}

= \frac{1}{4} + \frac{3}{4} \cos (4 x) + \frac{1}{4} {\frac{1}{4} - \frac{1}{2} \cos (4 x) + \frac{1}{4} \cos^{2} (4 x)}

= \frac{5}{16} + \frac{5}{8} \cos (4 x) + \frac{1}{16} (\frac{1}{2} + \frac{1}{2} \cos (8 x))

= \frac{11}{32} + \frac{5}{8} \cos (4 x) + \frac{1}{32} \cos (8 x)

Commented by john santu last updated on 28/Jan/20

\underset{0}{\int^{π}} (\frac{11}{32} + \frac{5}{8} \cos (4 x) + \frac{1}{32} \cos (8 x)) dx =

\frac{11 x}{32} + \frac{5}{32} \sin (4 x) + \frac{1}{256} \sin (8 x) \underset{0}{\overset{π}{∣}}

= \frac{11 π}{32} ★

Answered by Henri Boucatchou last updated on 28/Jan/20

w e h a v e {c o s}^{8} x + {s i n}^{8} x = 1 - \frac{8}{8} {s i n}^{2} 2 x

= {c o s}^{2} 2 x = \frac{1}{4} {(e^{i 2 x} + e^{- i 2 x})}^{2}

= \frac{1}{4} [(e^{i 4 x} + e^{- i 4 x}) + 2] = \frac{1}{4} (2 c o s 4 x + 2)

\Rightarrow \int_{0}^{π} ({c o s}^{8} x + {s i n}^{8} x) d x = \frac{1}{2} {[\frac{1}{4} s i n 4 x + 2 x]}_{0}^{π}

= \frac{1}{2} (2 π)

= π

Commented by john santu last updated on 28/Jan/20

how to get \sin^{8} x + \cos^{8} x = 1 - \frac{8}{8} \sin^{2} 2 x ?

it impossible

Commented by john santu last updated on 28/Jan/20

a^{4} + b^{4} = {(a + b)}^{4} - 2 ab ({2 a}^{2} - 3 ab - {2 b}^{2})

put a = \cos^{2} x, b = \sin^{2} x

\cos^{8} x + \sin^{8} x = 1 - \frac{1}{2} . 4 \sin^{2} xcos^{2} x (2 (\cos^{2} x - \sin^{2} x) - \frac{3}{4} . 4 \sin^{2} xcos^{2} x)

= 1 - \frac{1}{2} \sin^{2} (2 x) (2 \cos (2 x) - \frac{3}{4} \sin^{2} (2 x))

= 1 - \frac{1}{4} \sin (4 x) \sin (2 x) + \frac{3}{8} \sin^{4} (2 x)