∫ sin^8 x dx =?


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Question Number 160645 by tounghoungko last updated on 03/Dec/21

$\int \sin^{8} x dx = ?$
Commented by swamy1234 last updated on 04/Dec/21
good
	Answered by puissant last updated on 04/Dec/21
	$Ω=∫sin^8 x dx sin^8 x = (((e^(ix) −e^(−ix) )/(2i)))^8 = (1/(256))(e^(ix) −e^(−ix) )^8 = (1/(256)){e^(8ix) −8e^(7ix) e^(−ix) +28e^(6ix) e^(−2ix) −56e^(5ix) e^(−3ix) +70e^(i4x) e^(−i4x) −56e^(i3x) e^(−i5x) +28e^(i2x) e^(−i6x) −8e^(ix) e^(−i7x) +e^(−i8x) } = (1/(256)){(e^(i8x) +e^(−i8x) )−8(e^(i6x) +e^(−i6x) )+28(e^(i4x) +e^(−i4x) ) −56(e^(i2x) +e^(−2x) )+70} =(1/(128))cos8x−(1/(16))cos6x+(7/(32))cos4x−(7/(16))cos2x+((35)/(128)) Ω = ∫sin^8 x dx ; ⇒ Ω = (1/(984))cos8x−(1/(96))cos6x+(7/(128))cos4x−(7/(32))cos2x+((35x)/(128))+C (C∈R) ....................Le puissant.....................$
	$Ω = \int {s i n}^{8} x d x$ ${s i n}^{8} x = {(\frac{e^{i x} - e^{- i x}}{2 i})}^{8} = \frac{1}{256} {(e^{i x} - e^{- i x})}^{8}$ $= \frac{1}{256} {e^{8 i x} - 8 e^{7 i x} e^{- i x} + 28 e^{6 i x} e^{- 2 i x} - 56 e^{5 i x} e^{- 3 i x} + 70 e^{i 4 x} e^{- i 4 x}$ $- 56 e^{i 3 x} e^{- i 5 x} + 28 e^{i 2 x} e^{- i 6 x} - 8 e^{i x} e^{- i 7 x} + e^{- i 8 x}}$ $= \frac{1}{256} {(e^{i 8 x} + e^{- i 8 x}) - 8 (e^{i 6 x} + e^{- i 6 x}) + 28 (e^{i 4 x} + e^{- i 4 x})$ $- 56 (e^{i 2 x} + e^{- 2 x}) + 70}$ $= \frac{1}{128} c o s 8 x - \frac{1}{16} c o s 6 x + \frac{7}{32} c o s 4 x - \frac{7}{16} c o s 2 x + \frac{35}{128}$ $Ω = \int {s i n}^{8} x d x;$ $\Rightarrow Ω = \frac{1}{984} c o s 8 x - \frac{1}{96} c o s 6 x + \frac{7}{128} c o s 4 x - \frac{7}{32} c o s 2 x + \frac{35 x}{128} + C (C \in R)$ $. . . . . . . . . . . . . . . . . . . . L e p u i s s a n t . . . . . . . . . . . . . . . . . . . . .$
	Answered by peter frank last updated on 06/Dec/21

	$\sin^{n} xdx = - \frac{\sin^{n - 1} xcos x}{n} + \frac{n - 1}{n} \int \sin^{n - 2} xdx$ $\sin^{8} xdx = - \frac{\sin^{7} xcos x}{8} + \frac{7}{8} \int \sin^{6} xdx$ $\frac{7}{8} \int \sin^{6} xdx = . . . . (i)$ $- \frac{7 \sin^{5} xcos x}{48} + \frac{35}{64} \int \sin^{4} xdx$ $\frac{35}{64} \int \sin^{4} xdx . . . (ii)$ $- \frac{35 \sin^{3} xcos x}{256} + \frac{105}{256} \int \sin^{2} xdx$ $\frac{105}{256} \int \sin^{2} xdx . . . (iii)$ $- \frac{\sin xcos x}{2} + \frac{1}{2}$ $then substitute (i) (ii) (iii) then simplify$
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