Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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

Find ∫ Sin^6 x dx

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

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

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

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

complete it plzz

$${complete}\:{it}\:{plzz} \\ $$

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

pls go throuvh it

$${pls}\:{go}\:{throuvh}\:{it} \\ $$

Commented by Cheyboy last updated on 11/May/18

Thank you

$$\mathscr{T}{hank}\:{you} \\ $$$$ \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com