calculate ∫_0 ^π {cos^8 x +sin^8 x}dx


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Question Number 79763 by mathmax by abdo last updated on 28/Jan/20
$calculate ∫_0 ^π {cos^8 x +sin^8 x}dx$
$c a l c u l a t e \int_{0}^{π} {{c o s}^{8} x + {s i n}^{8} x} d x$
Commented by john santu last updated on 28/Jan/20
$cos^8 x+sin^8 x=(sin^4 x)^2 +(cos^4 x)^2 = (sin^4 x+cos^4 x)^2 −2sin^4 xcos^4 x = {(sin^2 x+cos^2 x)^2 −2sin^2 xcos^2 x}^2 −(1/2)sin^2 (2x) = {1−(1/2)sin^2 (2x)}^2 −(1/2)sin^2 (2x) =1−(3/2)sin^2 (2x)+(1/4)sin^4 (2x) =1−(3/2)((1/2)−(1/2)cos (4x))+(1/4){(1/2)−(1/2)cos (4x)}^2 =(1/4)+(3/4)cos (4x)+(1/4){(1/4)−(1/2)cos (4x)+(1/4)cos^2 (4x)} = (5/(16))+(5/8)cos (4x)+(1/(16))((1/2)+(1/2)cos (8x)) = ((11)/(32))+(5/8)cos (4x)+(1/(32))cos (8x)$
$\cos^{8} x + \sin^{8} x = {(\sin^{4} x)}^{2} + {(\cos^{4} x)}^{2}$ $= {(\sin^{4} x + \cos^{4} x)}^{2} - 2 \sin^{4} xcos^{4} x$ $= {{(\sin^{2} x + \cos^{2} x)}^{2} - 2 \sin^{2} xcos^{2} x}^{2} - \frac{1}{2} \sin^{2} (2 x)$ $= {1 - \frac{1}{2} \sin^{2} (2 x)}^{2} - \frac{1}{2} \sin^{2} (2 x)$ $= 1 - \frac{3}{2} \sin^{2} (2 x) + \frac{1}{4} \sin^{4} (2 x)$ $= 1 - \frac{3}{2} (\frac{1}{2} - \frac{1}{2} \cos (4 x)) + \frac{1}{4} {\frac{1}{2} - \frac{1}{2} \cos (4 x)}^{2}$ $= \frac{1}{4} + \frac{3}{4} \cos (4 x) + \frac{1}{4} {\frac{1}{4} - \frac{1}{2} \cos (4 x) + \frac{1}{4} \cos^{2} (4 x)}$ $= \frac{5}{16} + \frac{5}{8} \cos (4 x) + \frac{1}{16} (\frac{1}{2} + \frac{1}{2} \cos (8 x))$ $= \frac{11}{32} + \frac{5}{8} \cos (4 x) + \frac{1}{32} \cos (8 x)$
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

	$a^{4} + b^{4} = {(a + b)}^{4} - 2 ab ({2 a}^{2} - 3 ab - {2 b}^{2})$ $put a = \cos^{2} x, b = \sin^{2} x$ $\cos^{8} x + \sin^{8} x = 1 - \frac{1}{2} . 4 \sin^{2} xcos^{2} x (2 (\cos^{2} x - \sin^{2} x) - \frac{3}{4} . 4 \sin^{2} xcos^{2} x)$ $= 1 - \frac{1}{2} \sin^{2} (2 x) (2 \cos (2 x) - \frac{3}{4} \sin^{2} (2 x))$ $= 1 - \frac{1}{4} \sin (4 x) \sin (2 x) + \frac{3}{8} \sin^{4} (2 x)$
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