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Question Number 127250 by joki last updated on 28/Dec/20

∫sin^4 x dx

$$\int\mathrm{sin}^{\mathrm{4}} \mathrm{x}\:\mathrm{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 28/Dec/20

∫sin^2 x(1−cos^2 x)dx =∫sin^2 x−sin^2 xcos^2 x dx =∫(1/2)−(1/2)cos2x−(1/4)sin^2 2x=(x/2)−(1/4)sin2x−(1/8)∫1−cos4x dx =((3x)/8)−(1/4)sin2x+(1/(32))sin4x+C

$$\int{sin}^{\mathrm{2}} {x}\left(\mathrm{1}−{cos}^{\mathrm{2}} {x}\right){dx} \\ $$$$=\int{sin}^{\mathrm{2}} {x}−{sin}^{\mathrm{2}} {xcos}^{\mathrm{2}} {x}\:{dx} \\ $$$$=\int\frac{\mathrm{1}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{2}}{cos}\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{4}}{sin}^{\mathrm{2}} \mathrm{2}{x}=\frac{{x}}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{2}{x}−\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{1}−{cos}\mathrm{4}{x}\:{dx} \\ $$$$=\frac{\mathrm{3}{x}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}{sin}\mathrm{2}{x}+\frac{\mathrm{1}}{\mathrm{32}}{sin}\mathrm{4}{x}+{C} \\ $$

Answered by mathmax by abdo last updated on 28/Dec/20

∫ sin^4 xdx =∫(((1−cos(2x))/2))^2 dx =(1/4)∫(1−2cos(2x)+cos^2 (2x))dx =(x/4)−(1/4)sin(2x)+(1/8)∫(1+cos(4x))dx =(x/4)−(1/4)sin(2x)+(x/8) +(1/(32))sin(4x) +C =((3x)/8)−(1/4)sin(2x)+(1/(32))sin(4x) +C

$$\int\:\:\mathrm{sin}^{\mathrm{4}} \mathrm{xdx}\:=\int\left(\frac{\mathrm{1}−\mathrm{cos}\left(\mathrm{2x}\right)}{\mathrm{2}}\right)^{\mathrm{2}} \mathrm{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left(\mathrm{1}−\mathrm{2cos}\left(\mathrm{2x}\right)+\mathrm{cos}^{\mathrm{2}} \left(\mathrm{2x}\right)\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)+\frac{\mathrm{1}}{\mathrm{8}}\int\left(\mathrm{1}+\mathrm{cos}\left(\mathrm{4x}\right)\right)\mathrm{dx} \\ $$$$=\frac{\mathrm{x}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)+\frac{\mathrm{x}}{\mathrm{8}}\:+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\left(\mathrm{4x}\right)\:+\mathrm{C} \\ $$$$=\frac{\mathrm{3x}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\left(\mathrm{2x}\right)+\frac{\mathrm{1}}{\mathrm{32}}\mathrm{sin}\left(\mathrm{4x}\right)\:+\mathrm{C} \\ $$

Answered by physicstutes last updated on 28/Dec/20

let z = cos θ + i sin θ ⇒ (z − (1/z))^4 = (2i sin θ)^4 = 16 sin^4 θ ⇒ 16 sin^4 θ = z^4 + 4(z^3 )(−(1/z)) + 6(z^2 )(−(1/z))^2 + 4(z)(−(1/z))^3 + (−(1/z))^4 = (z^4 +(1/z^4 )) −4(z^2 +(1/z^2 ))−6 = 2 cos 4θ −8 cos 2θ −6 ⇒ sin^4 x = (1/(16))(2 cos 4x − 8cos 2x −6) ∫sin^4 x dx = (1/(16))∫(2cos 4x − 8cos 2x−6)dx = (1/(16))[((2 sin 4x)/4) − ((8 sin 2x)/2)−6x] + k ⇒ ∫sin^4 x dx = ((3x)/8)−(1/4)sin 2x + (1/(32)) sin 4x + k

$$\:\mathrm{let}\:{z}\:=\:\mathrm{cos}\:\theta\:+\:{i}\:\mathrm{sin}\:\theta \\ $$$$\Rightarrow\:\left({z}\:−\:\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \:=\:\left(\mathrm{2}{i}\:\mathrm{sin}\:\theta\right)^{\mathrm{4}} \:=\:\mathrm{16}\:\mathrm{sin}\:^{\mathrm{4}} \theta \\ $$$$\Rightarrow\:\:\mathrm{16}\:\mathrm{sin}^{\mathrm{4}} \theta\:=\:{z}^{\mathrm{4}} +\:\mathrm{4}\left({z}^{\mathrm{3}} \right)\left(−\frac{\mathrm{1}}{{z}}\right)\:+\:\mathrm{6}\left({z}^{\mathrm{2}} \right)\left(−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{2}} \:+\:\mathrm{4}\left({z}\right)\left(−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{3}} +\:\left(−\frac{\mathrm{1}}{{z}}\right)^{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left({z}^{\mathrm{4}} +\frac{\mathrm{1}}{{z}^{\mathrm{4}} }\right)\:−\mathrm{4}\left({z}^{\mathrm{2}} +\frac{\mathrm{1}}{{z}^{\mathrm{2}} }\right)−\mathrm{6} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}\:\mathrm{cos}\:\mathrm{4}\theta\:−\mathrm{8}\:\mathrm{cos}\:\mathrm{2}\theta\:−\mathrm{6} \\ $$$$\Rightarrow\:\mathrm{sin}^{\mathrm{4}} {x}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\left(\mathrm{2}\:\mathrm{cos}\:\mathrm{4}{x}\:−\:\mathrm{8cos}\:\mathrm{2}{x}\:−\mathrm{6}\right) \\ $$$$\:\int\mathrm{sin}^{\mathrm{4}} {x}\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\int\left(\mathrm{2cos}\:\mathrm{4}{x}\:−\:\mathrm{8cos}\:\mathrm{2}{x}−\mathrm{6}\right){dx}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\left[\frac{\mathrm{2}\:\mathrm{sin}\:\mathrm{4}{x}}{\mathrm{4}}\:−\:\frac{\mathrm{8}\:\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}}−\mathrm{6}{x}\right]\:+\:{k} \\ $$$$\:\Rightarrow\:\int\mathrm{sin}^{\mathrm{4}} {x}\:{dx}\:=\:\frac{\mathrm{3}{x}}{\mathrm{8}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{1}}{\mathrm{32}}\:\mathrm{sin}\:\mathrm{4}{x}\:+\:{k} \\ $$

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