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Integrate-0-1-Sin-2-2-x-dx-




Question Number 190172 by jlewis last updated on 29/Mar/23
Integrate   ∫_0 ^1  Sin^2 (2Πx)dx
$${Integrate}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \:{Sin}^{\mathrm{2}} \left(\mathrm{2}\Pi{x}\right){dx} \\ $$
Answered by mehdee42 last updated on 29/Mar/23
(1/2)∫_0 ^1 (1−cos2πx)dx=(1/2)(x−(1/(2π))sin2πx)_0 ^1 =(1/2)
$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\mathrm{1}−{cos}\mathrm{2}\pi{x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}\left({x}−\frac{\mathrm{1}}{\mathrm{2}\pi}{sin}\mathrm{2}\pi{x}\right)_{\mathrm{0}} ^{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by mr W last updated on 30/Mar/23
sin^2  (2πx)=((1−cos (4πx))/2)
$$\mathrm{sin}^{\mathrm{2}} \:\left(\mathrm{2}\pi{x}\right)=\frac{\mathrm{1}−\mathrm{cos}\:\left(\mathrm{4}\pi{x}\right)}{\mathrm{2}} \\ $$
Commented by mehdee42 last updated on 31/Mar/23
thanks for you
$${thanks}\:{for}\:{you} \\ $$

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