Find-Sin-6-x-dx-

Question Number 34827 by Cheyboy last updated on 11/May/18

$$\boldsymbol{{Find}}\:\int\:\boldsymbol{{Sin}}^{\mathrm{6}} \boldsymbol{{x}}\:\boldsymbol{{dx}} \\ $$$$ \\ $$

Commented by rahul 19 last updated on 11/May/18

sir how sin^6 xdx= (((e^(ix) −e^(−ix) )/(2i)))^6 ?

$${sir}\:{how}\:\:\:\mathrm{sin}\:^{\mathrm{6}} {xdx}=\:\left(\frac{{e}^{{ix}} −{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{\mathrm{6}} ? \\ $$

Commented by Cheyboy last updated on 11/May/18

$${Sir}\:{i}\:{dnt}\:{understand}\:{ur}\:{method} \\ $$$${plz}\:{be}\:{simplistic} \\ $$

Commented by abdo mathsup 649 cc last updated on 11/May/18

i have given the key becsuse ∫ e^(i(2k−6)x) dx = (1/(2k−6)) e^(i(2k−6)x) +λ =(1/(2k−6))( cos(3k−6)x +isin(3k−6)x) +λ.....

$${i}\:{have}\:{given}\:{the}\:{key}\:{becsuse} \\ $$$$\int\:\:{e}^{{i}\left(\mathrm{2}{k}−\mathrm{6}\right){x}} {dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{6}}\:{e}^{{i}\left(\mathrm{2}{k}−\mathrm{6}\right){x}} \:+\lambda \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{k}−\mathrm{6}}\left(\:{cos}\left(\mathrm{3}{k}−\mathrm{6}\right){x}\:+{isin}\left(\mathrm{3}{k}−\mathrm{6}\right){x}\right)\:+\lambda….. \\ $$

Commented by math khazana by abdo last updated on 11/May/18

another method (linearization) ∫ sin^6 xdx = ∫ ( ((e^(ix) −e^(−ix) )/(2i)))^6 dx = (1/((2i)^6 )) ∫ Σ_(k=0) ^6 C_6 ^k e^(ikx) e^(−i(6−k)x) dx = (1/((2i)^6 )) Σ_(k=0) ^6 C_6 ^k ∫ e^(i(2k−6)x) dx ....

$${another}\:{method}\:\:\left({linearization}\right) \\ $$$$\int\:{sin}^{\mathrm{6}} {xdx}\:\:=\:\int\:\:\:\left(\:\:\frac{{e}^{{ix}} \:\:\:−{e}^{−{ix}} }{\mathrm{2}{i}}\right)^{\mathrm{6}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{6}} }\:\:\int\:\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \:{C}_{\mathrm{6}} ^{{k}} \:\:\:\:{e}^{{ikx}} \:\:{e}^{−{i}\left(\mathrm{6}−{k}\right){x}} \:{dx} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{\mathrm{6}} }\:\:\sum_{{k}=\mathrm{0}} ^{\mathrm{6}} \:{C}_{\mathrm{6}} ^{{k}} \:\:\:\:\:\int\:\:\:{e}^{{i}\left(\mathrm{2}{k}−\mathrm{6}\right){x}} {dx}\:…. \\ $$

Commented by abdo mathsup 649 cc last updated on 11/May/18

its a formula ((e^(ix) −e^(−ix) )/(2i)) =((2iIm(e^(ix) ))/(2i)) = sinx .

$${its}\:{a}\:{formula}\:\:\:\frac{{e}^{{ix}} \:\:−{e}^{−{ix}} }{\mathrm{2}{i}}\:=\frac{\mathrm{2}{iIm}\left({e}^{{ix}} \right)}{\mathrm{2}{i}} \\ $$$$=\:{sinx}\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/May/18

=∫(((1−cos2x)^3 )/2^3 ) (1/8)∫((1−3cos2x+3cos^2 2x−cos^3 2x)/) (1/8)∫dx−(3/8)∫cos2x dx+(3/8)∫(((1+cos4x))/2)−(1/8)∫cos^3 2x =(x/8)−(3/(16))sin2x+(3/(16))(x+((sin4x)/4))−(1/8)∫((cos6x+3cos2x)/4) =do−(1/(32))∫cos6x+3cos2x =do−(1/(32)) ×((sin6x)/6)−(3/(32))×((sin2x)/2) =(x/8)−(3/(16))sin2x+(3/(16))(x+((sin4x)/4))−(1/(32))((sin6x)/6)−(3/(64))× sin2x x((1/8)+(3/(16)))−sin2x((3/(16))+(3/(64)))+sin4x((3/(64))) −sin6x((1/(192))) =x((5/(16)))−sin2x(((15)/(64)))+sin4x((3/(64)))−(1/(192))sin6x =(5/(16))x−((15)/(64))sin2x+(3/(64))sin4x−(1/(192))sin6x

$$=\int\frac{\left(\mathrm{1}−{cos}\mathrm{2}{x}\right)^{\mathrm{3}} }{\mathrm{2}^{\mathrm{3}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{1}−\mathrm{3}{cos}\mathrm{2}{x}+\mathrm{3}{cos}^{\mathrm{2}} \mathrm{2}{x}−{cos}^{\mathrm{3}} \mathrm{2}{x}}{} \\ $$$$\frac{\mathrm{1}}{\mathrm{8}}\int{dx}−\frac{\mathrm{3}}{\mathrm{8}}\int{cos}\mathrm{2}{x}\:{dx}+\frac{\mathrm{3}}{\mathrm{8}}\int\frac{\left(\mathrm{1}+{cos}\mathrm{4}{x}\right)}{\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{8}}\int{cos}^{\mathrm{3}} \mathrm{2}{x} \\ $$$$=\frac{{x}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{16}}{sin}\mathrm{2}{x}+\frac{\mathrm{3}}{\mathrm{16}}\left({x}+\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{8}}\int\frac{{cos}\mathrm{6}{x}+\mathrm{3}{cos}\mathrm{2}{x}}{\mathrm{4}} \\ $$$$={do}−\frac{\mathrm{1}}{\mathrm{32}}\int{cos}\mathrm{6}{x}+\mathrm{3}{cos}\mathrm{2}{x} \\ $$$$={do}−\frac{\mathrm{1}}{\mathrm{32}}\:×\frac{{sin}\mathrm{6}{x}}{\mathrm{6}}−\frac{\mathrm{3}}{\mathrm{32}}×\frac{{sin}\mathrm{2}{x}}{\mathrm{2}} \\ $$$$=\frac{{x}}{\mathrm{8}}−\frac{\mathrm{3}}{\mathrm{16}}{sin}\mathrm{2}{x}+\frac{\mathrm{3}}{\mathrm{16}}\left({x}+\frac{{sin}\mathrm{4}{x}}{\mathrm{4}}\right)−\frac{\mathrm{1}}{\mathrm{32}}\frac{{sin}\mathrm{6}{x}}{\mathrm{6}}−\frac{\mathrm{3}}{\mathrm{64}}× \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{sin}\mathrm{2}{x} \\ $$$${x}\left(\frac{\mathrm{1}}{\mathrm{8}}+\frac{\mathrm{3}}{\mathrm{16}}\right)−{sin}\mathrm{2}{x}\left(\frac{\mathrm{3}}{\mathrm{16}}+\frac{\mathrm{3}}{\mathrm{64}}\right)+{sin}\mathrm{4}{x}\left(\frac{\mathrm{3}}{\mathrm{64}}\right) \\ $$$$−\mathrm{sin6x}\left(\frac{\mathrm{1}}{\mathrm{192}}\right) \\ $$$$=\mathrm{x}\left(\frac{\mathrm{5}}{\mathrm{16}}\right)−\mathrm{sin2x}\left(\frac{\mathrm{15}}{\mathrm{64}}\right)+\mathrm{sin4x}\left(\frac{\mathrm{3}}{\mathrm{64}}\right)−\frac{\mathrm{1}}{\mathrm{192}}\mathrm{sin6x} \\ $$$$=\frac{\mathrm{5}}{\mathrm{16}}\mathrm{x}−\frac{\mathrm{15}}{\mathrm{64}}\mathrm{sin2x}+\frac{\mathrm{3}}{\mathrm{64}}\mathrm{sin4x}−\frac{\mathrm{1}}{\mathrm{192}}\mathrm{sin6x} \\ $$$$\:\:\:\: \\ $$$$ \\ $$

Commented by Cheyboy last updated on 11/May/18