What is the integral `int sin^6 x dx` 

1 Answer | Add Yours

Top Answer

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

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

`int sin^6 x dx`

=> `int (sin^2 x)^3 dx`

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

=> `int -cos^6 x + 3*cos^4x - 3*cos^2 x + 1 dx`

Now use the identity `cos 2x = 2*cos^2x - 1` or `cos^2x = (1/2)*(1 + cos 2x)` repeatedly

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

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

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

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

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

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

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

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

=>

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

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

`int 5/16 - (3/8)*cos 2x + (3/16)*cos 4x dx`

=> `(5/16)*x - (3/16)*sin 2x + (3/64)*sin 4x`

To determine `int (1 - sin^2 2x)*cos 2x dx` substiitute `y = sin 2x` , `(1/2)*dy = cos 2x dx`

=> `(1/2)*int 1 - y^2 dy`

=> `(1/2)*y - (1/6)*y^3`

substituting y = sin 2x

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

Adding the two gives:

`(5/16)*x - (3/16)*sin 2x + (3/64)*sin 4x - (1/16)*sin 2x + (1/48)*sin^3 2x`

=> `(60x - 48*sin 2x + 4*sin^3 2x + 9*sin 4x)/192`

The integral `int sin^6 x dx` = `(60x - 48*sin 2x + 4*sin^3 2x + 9*sin 4x)/192`

We’ve answered 318,991 questions. We can answer yours, too.

Ask a question