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please explain how to evaluate the indefinite integral `int sin^3(6theta)d theta`...

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user7230927 | Salutatorian

Posted August 5, 2013 at 6:07 PM via web

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please explain how to evaluate the indefinite integral `int sin^3(6theta)d theta`  using u-substitution

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 5, 2013 at 6:22 PM (Answer #1)

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You need to evaluate the indefinite integral, using substitution, such that:

`cos(6theta) = u => -6sin(6theta)d theta = du => sin(6theta)d theta = -(du)/6`

`int sin^3(6theta)d theta = int sin^2(6theta)*sin(6theta)d theta`

Using the basic formula of trigonometry, yields:

`sin^2(6theta) = 1 - cos^2(6theta)`

`int sin^2(6theta)*sin(6theta)d theta = int (1 - cos^2(6theta))sin(6theta)d theta`

Using the property of linearity of integral, yields:

`int sin^2(6theta)*sin(6theta)d theta = int sin(6theta)d theta - int cos^2(6theta)sin(6theta)d theta`

`int sin^2(6theta)*sin(6theta)d theta = -(cos(6theta))/6 - int cos^2(6theta)sin(6theta)d theta`

Changing the variable in ` int cos^2(6theta)sin(6theta)d theta` yields:

`int cos^2(6theta)sin(6theta)d theta = - int u^2*(du)/6 = -u^3/18 `

Replacing back `cos(6theta)` for u, yields:

`int cos^2(6theta)sin(6theta)d theta = - (cos^3(6theta))/18 + c`

`int sin^3(6theta)d theta = (cos^3(6theta))/18 - (cos(6theta))/6 + c`

Factoring out `(cos(6theta))/6` yields:

`int sin^3(6theta)d theta = (cos(6theta))/6((cos^2(6theta))/3 - 1) + c`

Hence, evaluating the indefinite integral, using the indicated substitution, yields `int sin^3(6theta)d theta = (cos(6theta))/6((cos^2(6theta))/3 - 1) + c.`

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mariloucortez | High School Teacher | (Level 3) Adjunct Educator

Posted August 5, 2013 at 7:05 PM (Answer #2)

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`int sin^3(6theta) d theta`

By u-substitution, you let

u = `6 theta`

du = `6d theta`

`(du)/6 = d theta`

Substitute the derived values in the original problem

`intsin^3(u) (du)/6`

Using trigonometric identity `sin^2(x) = 1-cos^2(x).`

`intsin^3(u) (du/6) = intsin^2(u)sin(u)(du)/6` 
`(1/6)int(1-cos^2(u))sin(u) du`

Distribute `sin(u)du`  into the ().

`1/6[intsin(u)du - intcos^2(u)sin(u)du]`

For the first integral, it is directly integrable with the formula:

`intsin(u) du = -cos(u) + c`

For the second integral you can apply power formula:

`intx^n dx = (x^(n+1))/(n+1)`

`x = cos(u)`

`n = 2` 
`dx = sin(u) du` 

Notice that to use this, `dx`  is the derivative of `x` . Thus, `sin(u)du`  is the derivative of `cos(u).`

So for the second integral, you have:

`intcos^2(u)sin(u)du = (cos^3(u))/3 + c`

Combining the answers, you have:

`1/6[-cos(u) + (cos^3(u))/3 + c]`

* constant plus constant is equal to constant, c + c = C or constant multiplied with constant is always a constant...

Transform back that answer by substituting `u = 6 theta`

`1/6[-cos(6theta) + (cos^3(6theta))/3] + c`

`=(cos(6 theta))/6[(cos^2(6 theta))/3 - 1] + c`

Thus, the indefinite integral is `(cos^3(6theta))/18 - cos(6theta)/6 + c`

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