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Question Number 36128 by mondodotto@gmail.com last updated on 29/May/18

∫sin^8 xdx ∫sin^6 xdx

$$\int\boldsymbol{\mathrm{sin}}^{\mathrm{8}} \boldsymbol{{xdx}} \\ $$$$\int\boldsymbol{\mathrm{sin}}^{\mathrm{6}} \boldsymbol{{xdx}} \\ $$

Commented by Joel579 last updated on 29/May/18

Commented by abdo mathsup 649 cc last updated on 29/May/18

∫ sin^8 xdx = ∫( ((1−cos(2x))/2))^4 dx = (1/(32)) ∫ (cos^2 (2x)−2cos(2x)+1)^2 dx = (1/(32))∫ ( ((1+cos(2x))/2) −2cos(2x) +1)^2 dx = (1/(64)) ∫ (1+cos(2x)−4cos(2x) +2)^2 dx =(1/(64)) ∫ (3 −3 cos(2x))^2 dx =(9/(64)) ∫ (cos^2 (2x) −2cos(2x) +1)dx =(9/(64))∫ ( ((1+cos(4x))/2) −2 cos(2x) +1)dx = (9/(128)) ∫ { 1+cos(4x) −4cos(2x) +2}dx =(9/(128))∫ { cos(4x) −4 cos(2x) +3}dx = (9/(4.128)) sin(4x) −((18)/(128))sin(2x) +((27)/(128)) +c

$$\int\:{sin}^{\mathrm{8}} {xdx}\:=\:\int\left(\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right)^{\mathrm{4}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{32}}\:\int\:\:\:\left({cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)+\mathrm{1}\right)^{\mathrm{2}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{32}}\int\:\left(\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\:−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{1}\right)^{\mathrm{2}} {dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{64}}\:\int\:\:\left(\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{2}\right)^{\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{64}}\:\int\:\left(\mathrm{3}\:−\mathrm{3}\:{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} {dx} \\ $$$$=\frac{\mathrm{9}}{\mathrm{64}}\:\int\:\:\left({cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\:−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{1}\right){dx} \\ $$$$=\frac{\mathrm{9}}{\mathrm{64}}\int\:\:\left(\:\:\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\:−\mathrm{2}\:{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{1}\right){dx} \\ $$$$=\:\frac{\mathrm{9}}{\mathrm{128}}\:\int\:\left\{\:\:\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)\:−\mathrm{4}{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{2}\right\}{dx} \\ $$$$=\frac{\mathrm{9}}{\mathrm{128}}\int\:\left\{\:{cos}\left(\mathrm{4}{x}\right)\:−\mathrm{4}\:{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{3}\right\}{dx} \\ $$$$=\:\:\:\frac{\mathrm{9}}{\mathrm{4}.\mathrm{128}}\:{sin}\left(\mathrm{4}{x}\right)\:−\frac{\mathrm{18}}{\mathrm{128}}{sin}\left(\mathrm{2}{x}\right)\:+\frac{\mathrm{27}}{\mathrm{128}}\:+{c}\: \\ $$$$ \\ $$

Commented by abdo mathsup 649 cc last updated on 29/May/18

∫ sin^6 xdx = ∫ (((1−cos(2x))/3))^3 dx =(1/(27))∫ (1−cos(2x))^2 (1−cos(2x))dx = (1/(27)) ∫ ( 1−2cos(2x) +cos^2 (2x))(1−cos(2x))dx = (1/(27)) ∫ ( 1−cos(2x) +((1+cos(4x))/2))(1−cos(2x))dx = (1/(54)) ∫ ( 3−2cos(2x) +cos(4x))(1−cos(2x))dx = (1/(54)) ∫ (3−2cos(2x) +cos(4x))dx −(1/(54)) ∫ (3−2cos(2x) +cos(4x))cos(2x)dx = (3/(54))x −(1/(54)) sin(2x) +(1/(216)) sin(4x) −(3/(54)) ∫ cos(2x)dx +(2/(54)) ∫ cos^2 (2x)dx −(1/(54)) ∫ cos(2x)cos(4x)dx =....

$$\int\:{sin}^{\mathrm{6}} {xdx}\:=\:\int\:\left(\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{3}}\right)^{\mathrm{3}} {dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{27}}\int\:\:\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)^{\mathrm{2}} \left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{27}}\:\int\:\:\left(\:\mathrm{1}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right)\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{27}}\:\int\:\left(\:\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\:+\frac{\mathrm{1}+{cos}\left(\mathrm{4}{x}\right)}{\mathrm{2}}\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{54}}\:\int\:\:\left(\:\mathrm{3}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+{cos}\left(\mathrm{4}{x}\right)\right)\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right){dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{54}}\:\int\:\:\left(\mathrm{3}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+{cos}\left(\mathrm{4}{x}\right)\right){dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{54}}\:\:\int\:\:\left(\mathrm{3}−\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+{cos}\left(\mathrm{4}{x}\right)\right){cos}\left(\mathrm{2}{x}\right){dx} \\ $$$$=\:\frac{\mathrm{3}}{\mathrm{54}}{x}\:\:−\frac{\mathrm{1}}{\mathrm{54}}\:{sin}\left(\mathrm{2}{x}\right)\:+\frac{\mathrm{1}}{\mathrm{216}}\:{sin}\left(\mathrm{4}{x}\right) \\ $$$$−\frac{\mathrm{3}}{\mathrm{54}}\:\int\:{cos}\left(\mathrm{2}{x}\right){dx}\:\:+\frac{\mathrm{2}}{\mathrm{54}}\:\int\:{cos}^{\mathrm{2}} \left(\mathrm{2}{x}\right){dx} \\ $$$$−\frac{\mathrm{1}}{\mathrm{54}}\:\int\:\:\:{cos}\left(\mathrm{2}{x}\right){cos}\left(\mathrm{4}{x}\right){dx} \\ $$$$=.... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 29/May/18

Answered by tanmay.chaudhury50@gmail.com last updated on 29/May/18

Answered by tanmay.chaudhury50@gmail.com last updated on 29/May/18

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