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

∫ sin^8 x dx =?

$$\:\int\:\mathrm{sin}\:^{\mathrm{8}} \mathrm{x}\:\mathrm{dx}\:=? \\ $$

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.....................

$$\Omega=\int{sin}^{\mathrm{8}} {x}\:{dx}\:\: \\ $$$${sin}^{\mathrm{8}} {x}\:=\:\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{\mathrm{8}} =\:\frac{\mathrm{1}}{\mathrm{256}}\left({e}^{{ix}} −{e}^{−{ix}} \right)^{\mathrm{8}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{256}}\left\{{e}^{\mathrm{8}{ix}} −\mathrm{8}{e}^{\mathrm{7}{ix}} {e}^{−{ix}} +\mathrm{28}{e}^{\mathrm{6}{ix}} {e}^{−\mathrm{2}{ix}} −\mathrm{56}{e}^{\mathrm{5}{ix}} {e}^{−\mathrm{3}{ix}} +\mathrm{70}{e}^{{i}\mathrm{4}{x}} {e}^{−{i}\mathrm{4}{x}} \right. \\ $$$$\left.−\mathrm{56}{e}^{{i}\mathrm{3}{x}} {e}^{−{i}\mathrm{5}{x}} +\mathrm{28}{e}^{{i}\mathrm{2}{x}} {e}^{−{i}\mathrm{6}{x}} −\mathrm{8}{e}^{{ix}} {e}^{−{i}\mathrm{7}{x}} +{e}^{−{i}\mathrm{8}{x}} \right\} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{256}}\left\{\left({e}^{{i}\mathrm{8}{x}} +{e}^{−{i}\mathrm{8}{x}} \right)−\mathrm{8}\left({e}^{{i}\mathrm{6}{x}} +{e}^{−{i}\mathrm{6}{x}} \right)+\mathrm{28}\left({e}^{{i}\mathrm{4}{x}} +{e}^{−{i}\mathrm{4}{x}} \right)\right. \\ $$$$\left.−\mathrm{56}\left({e}^{{i}\mathrm{2}{x}} +{e}^{−\mathrm{2}{x}} \right)+\mathrm{70}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{128}}{cos}\mathrm{8}{x}−\frac{\mathrm{1}}{\mathrm{16}}{cos}\mathrm{6}{x}+\frac{\mathrm{7}}{\mathrm{32}}{cos}\mathrm{4}{x}−\frac{\mathrm{7}}{\mathrm{16}}{cos}\mathrm{2}{x}+\frac{\mathrm{35}}{\mathrm{128}} \\ $$$$\Omega\:=\:\int{sin}^{\mathrm{8}} {x}\:{dx}\:; \\ $$$$\Rightarrow\:\Omega\:=\:\frac{\mathrm{1}}{\mathrm{984}}{cos}\mathrm{8}{x}−\frac{\mathrm{1}}{\mathrm{96}}{cos}\mathrm{6}{x}+\frac{\mathrm{7}}{\mathrm{128}}{cos}\mathrm{4}{x}−\frac{\mathrm{7}}{\mathrm{32}}{cos}\mathrm{2}{x}+\frac{\mathrm{35}{x}}{\mathrm{128}}+{C}\:\left({C}\in\mathbb{R}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:....................\mathscr{L}{e}\:{puissant}..................... \\ $$

Answered by peter frank last updated on 06/Dec/21

sin^n xdx=−((sin^(n−1) xcos x)/n)+((n−1)/n)∫sin^(n−2) xdx sin^8 xdx=−((sin^7 xcos x)/8)+(7/8)∫sin^6 xdx (7/8)∫sin^6 xdx=....(i) −((7sin^5 xcos x)/(48))+((35)/(64))∫sin^4 xdx ((35)/(64))∫sin^4 xdx...(ii) −((35sin^3 xcos x)/(256))+((105)/(256))∫sin^2 xdx ((105)/(256))∫sin^2 xdx...(iii) −((sin xcos x)/2)+(1/2) then substitute (i) (ii)(iii) then simplify

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

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