Homework Help

Find the integral `int sin^6x dx`

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soccerfan55 | Student | (Level 1) Honors

Posted October 8, 2012 at 2:36 PM via web

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Find the integral `int sin^6x dx`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted October 9, 2012 at 4:58 AM (Answer #1)

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You should use the half angle identity to solve the integral such that:

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

Reasoning by analogy yields:

`sin^6 2x = (sin^2 x)(sin^2 x)(sin^2 x)`

`sin^6 2x = ((1 - cos 2x)/2)((1 - cos 2x)/2)((1 - cos 2x)/2)`

`sin^6 2x = (1/8)((1 - cos 2x)^3)`

Expanding the cube yields:

`sin^6 2x = (1/8)(1 - 3cos 2x + 3cos^2 2x - cos^3 2x)`

Hence, evaluating the integral yields:

`int sin^6 x dx= int (1/8)(1 - 3cos x + 3cos^2 x - cos^3 x) dx`

Using the linearity yields:

`int sin^6 2x dx = (1/8) int dx - (3/8)int cos 2x dx +(3/8)int cos^2 2x dx - (1/8) int cos^3 2x dx`

You should use the half angle identity to solve `int cos^2 2x dx`  and  `int cos^3 2x dx`  such that:

`int cos^2 2x dx = int(1 + cos4x)/2 dx`

Using the linearity yields:

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

`int cos^2 2x dx = x/2 + (sin 4x)/8 + c`

You need to solve `int cos^3 2x dx`  such that:

`int cos^3 2x dx = int cos^2 2x*cos 2xdx`

Using the fundamental formula of trigonometry yields:

`int cos^3 2x dx = int (1- sin^2 2x)*cos 2x dx`

`int cos^3 2x dx = int dx - int sin^2 2x*cos 2x dx`

You need to solve `int sin^2 2x*cos 2x dx`  using substitution such that:

`sin 2x = t => 2cos 2x dx = dt => cos 2x dx = (dt)/2`

`int sin^2 2x*cos 2x dx = int t^2*(dt)/2`

`int sin^2 2x*cos 2x dx = (1/2)int t^2 dt`

`int sin^2 2x*cos 2x dx = (sin^3 2x)/6 + c`

`int cos^3 2x dx = x - (sin^3 2x)/6 + c`

`int sin^6 2x dx = (1/8) x - (3sin 2x)/16 + (3/8)(x/2 + (sin 4x)/8) -`  `(1/8)(x - (sin^3 2x)/6) + c `

`int sin^6 2x dx = (1/8) x - (3sin 2x)/16 + (3x)/16 + (3sin 4x)/64 - (1/8)x + (sin^3 2x)/48 + c`

Reducing like terms yields:

`int sin^6 2x dx = - (3sin 2x)/16 + (3x)/16 + (3sin 4x)/64 + (sin^3 2x)/48 + c`

Hence, evaluating the given integral yields `int sin^6 2x dx = - (3sin 2x)/16 + (3x)/16 + (3sin 4x)/64 + (sin^3 2x)/48 + c` .

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 9, 2012 at 9:33 AM (Answer #2)

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The integral `int sin^6x dx` has to be determined.

Use the rule `cos 2x = 1 - 2*sin^2x => sin^2x = (1 - cos 2x)/2`

 `sin^6x = (sin^2x)^3 = ((1 - cos 2x)/2)^3`

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

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

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

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

The integral `int sin^6x dx`

=> `(1/8)*int 1 - 3*cos 2x + 3/2 + (3*cos 4x)/2 - cos 2x + sin^2 2x*cos 2x dx`

=> `(1/8)*int 5/2 - 4*cos 2x + (3*cos 4x)/2 + sin^2 2x*cos 2x dx`

=> `(5x)/16 - (1/2)*int cos 2x dx +(3/16)*int cos 4x dx + (1/8)*int sin^2 2x*cos 2x dx`

=> `(5x)/16 - (sin 2x)/4 + (3*sin 4x)/64 + (1/8)*int sin^2 2x*cos2x dx`

To determine `(1/8)*int sin^2 2x*cos 2x dx`

let `sin 2x = y => dy = 2*cos 2x dx`

=> `(1/16)*int y^2 dx`

=> `y^3/48`

substitute `y = sin 2x`

=> `(sin^3 2x)/48`

The integral `int sin^6x dx = (5x)/16 - (sin 2x)/4 + (3*sin 4x)/64 + (sin^3 2x)/48` 

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