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evaluate the indefinite integral: S [ sin^2 x + 6x * (3x + 2 )^2 + cos^2 x] dx  

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bribribriw | Student, Undergraduate | eNotes Newbie

Posted November 22, 2011 at 5:07 AM via web

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evaluate the indefinite integral: S [ sin^2 x + 6x * (3x + 2 )^2 + cos^2 x] dx

 

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beckden | High School Teacher | (Level 1) Educator

Posted November 22, 2011 at 6:18 AM (Answer #1)

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int sin^2x + 6x *(3x+2)^2 + cos^2x dx

First notice that sin^2x + cos^2x = 1 so our integral is

int 1 + 6x * (3x + 2)^2 dx = x + int 6x(3x+2)^2 dx

Expanding 6x(3x+2)^2 = 6x(9x^2 + 12x + 4) = 54x^3 + 72x^2 + 24x

We get

int 1 + 6x(3x+2)^2 dx = int (1 + 54x^3 + 72x^2 + 24x) dx

= x + 54/4 x^4 + 72/3 x^3 + 24/2 x^2 + C

= 27/2 x^4 + 24x^3 + 12x^2 + x + C  is our answer

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted November 22, 2011 at 7:16 PM (Answer #2)

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You may also use the formula of half angle for sin^2 x and cos^2 x.

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

cos ^2 x= (1+cos(2x))/2

The integral of sin^2 x is:

`Int sin^2 x dx=Int (1-cos(2x))/2` dx`=(1/2)*(Int (dx)-Int (cos 2x)dx)`

`Int sin^2x dx=(1/2)*x - (sin 2x)/4 + C`

`Int cos^2 x dx=(1/2)*(Int (dx)+Int (cos 2x)dx)`

`Int (sin^2 x + cos ^2 x)dx = 2*(1/2)*x + (sin 2x)/4 - (sin 2x)/4 + C`

Reduce `(sin 2x)/4 and - (sin 2x)/4` :

`Int (sin^2 x + cos ^2 x)dx = x + C`

Calculate Int `6x*(3x+2)^2`  using substitution.

`3x+2=y =gt x=(y-2)/3`

Differentiate

`dx=(1/3)*dy`

`Int 6x*(3x+2)^2=Int 6*(y-2)*y^2dy/3=Int 2*y^2(y-2)`

`Int 2*y^2(y-2)=Int (2y^3-4y^2)dy=Int 2y^3dy -Int 4y^2dy `

`Int 2*y^2(y-2)=2y^4/4 - 4y^3/3+C`

`Int 6x*(3x+2)^2=(3x+2)^4/2 - 4(3x+2)^3/3+C`

The final result: `Int(sin^2 x+6x*(3x+2)^2+cos^2 x)dx=x + (3x+2)^4/2 - 4(3x+2)^3/3+C`

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