Menu Close

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
Integrate01Sin2(2Πx)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)
1201(1cos2πx)dx=12(x12πsin2πx)01=12
Commented by mr W last updated on 30/Mar/23
sin^2  (2πx)=((1−cos (4πx))/2)
sin2(2πx)=1cos(4πx)2
Commented by mehdee42 last updated on 31/Mar/23
thanks for you
thanksforyou

Leave a Reply

Your email address will not be published. Required fields are marked *