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find the indefinite integral of h(u) = sin^2 (8u)

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user1548464 | eNoter

Posted March 25, 2013 at 7:00 PM via web

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find the indefinite integral of h(u) = sin^2 (8u)

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted March 25, 2013 at 7:07 PM (Answer #1)

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You need to come up with the following substitution, such that:

`8u = v => (8u)' = v' => 8du = dv => du = (dv)/8`

Changing the vaiable, yields:

`int sin^2 (8u) du = int sin^2 v (dv)/8`

You need to use the half angle identity, such that:

`sin^2 v = (1 - cos 2v)/2`

`int sin^2 v*(dv)/8 = int (1 - cos 2v)/2*(dv)/8`

You need to use the property of linearity of indefinite integral, such that:

`int (1 - cos 2v)/16*(dv) = int (1/16)dv - int (cos 2v)/16 dv`

`int (1 - cos 2v)/16*(dv) = (1/16) v - (sin 2v)/32 + c`

Replacing back `8u` for `v` yields:

`intsin^2 (8u) du = (1/16) *(8u) - (sin 16u)/32 + c`

`intsin^2 (8u) du = u/2 - (sin 16u)/32 + c`

Hence, evaluating the indefinite integral, using substitution and half angle trigonometric identity, yields `intsin^2 (8u) du = u/2 - (sin 16u)/32 + c.`

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pramodpandey | College Teacher | Valedictorian

Posted March 26, 2013 at 6:28 AM (Answer #2)

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`cos(2t)=1-2sin^2(t)`

`2sin^2(t)=1-cos(2t)`

`sin^2(t)=(1-cos(2t))/2`

`sin^2(8u)=(1-cos(16u))/2`

`intsin^2(8u)du=int(1-cos(16u))/2du`

`=(1/2)(int1du-intcos(16u)du)`

`=(1/2)(u-sin(16u)/16)+c`

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