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`intsin^4(2x)dx`

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roshan-rox | Valedictorian

Posted August 2, 2013 at 4:08 AM via web

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`intsin^4(2x)dx`

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aruv | High School Teacher | Valedictorian

Posted August 2, 2013 at 4:23 AM (Answer #1)

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`sin^4(2x)=(sin^2(2x))^2=((1-cos(4x))/2)^2`

`=(1/4)(1+cos^2(4x)-2cos(4x))`

`=(1/4)(1-2cos(4x))+(1/4)(cos(8x)+1)/2`

`` `=(1/4)-(1/2)cos(4x)+(1/8)cos(8x)+1/8`

`=(3/8)-(1/2)cos(4x)+(1/8)cos(8x)`

Thus

`intsin^4(2x)dx=int((3/8)-(1/2)cos(4x)+(1/8)cos(8x))dx+c`

`=int3/8dx-intcos(4x)/2dx+intcos(8x)/8dx+c`

`=(3x)/8-sin(4x)/8+sin(8x)/64` +c

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