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What is the integral of `int sin^2(3x)*cos^2(3x) dx`

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lenalen | Student, Undergraduate | eNoter

Posted December 8, 2011 at 10:27 PM via web

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What is the integral of `int sin^2(3x)*cos^2(3x) dx`

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted December 8, 2011 at 11:13 PM (Answer #1)

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`int sin^2 (3x) cos^2 (3x) dx`

`` We know that:

`sin(3x)= 3sinx -4sin^3 x `

`cos(3x)= 4cos^3 x -3cosx `

`==gt int (3sinx-4sin^3 x )(4cos^3 x -3cosx) dx `

`==gt int (12sinxcos^3 x-9sinxcosx -16sin^3 x cos^3 x +12cosx sin^3 x) dx `

`==gt 12int sinxcos^3 x dx -9int sinxcosx dx`

`-16int sin^3 x cos^2 x dx+ 12int cosx sin^3 x dx`

`==gt 12int sinx cos^3 x dx `

`==gt u= cosx ==gt du= sinx dx `

`==gt 12int sinx u^3 du/sinx = 12int u^3 du = 12u^4/4= 3u^4 `

`==gt 12int sinxcos^3 x = 3cos^4 x.............(1) `

`9 int sinxcosx dx `

`==gt u= cosx ==gt du= sinx dx `

`==gt 9int sinx u du/sinx = 9int u du= 9u^2/2 = (9/2)u^2 `

`==gt 9int sinxcosx dx = (9/2)cos^2 x......(2)`

`16 int sin^3 x cos^3 x dx = 16int (sinxcosx)^3 dx= 16int (sin2x)^3/2 `

`==gt 8int sin^3 (2x) dx = 8int sin^2 (2x) sin(2x) `

`==gt 8int (1-cos^2 2x) sin2x) dx `

`==gt u= cos2x ==gt du= sin(2x)/2 dx==gt dx= (2du)/sin(2x) `

`==gt 8int (1-u^2) sin2x (2du)/sin(2x)= 16int (1-u^2) du `

`==gt 16u - 16u^3/3 `

`==gt 16int sin^3 x cos^3 x = 16cos2x -(16/3)cos^3 (2x)........(3) `

`12 int sin^3 cosx dx `

`==gt u= sinx ==gt du= -cosx dx `

`==gt -12int u^3 du = -12u^4/4 = -3u^4 `

`==gt 12int sin^3 x cosx = -3sin^4 x..........(4) `

`==gt int sin^2 (3x) cos^2 (3x)=`

`= 3cos^4 x +(9/2)cos^2 x+ 16cos(2x) -(16/3)cos^3 (2x)-3sin^4 x`

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