Question

Evaluate the integral? : `int 8cos^4 2pit dt`

Answer 1

`int \ 8cos^4 2pit \ dt = 3x + (sin(4pix))/(pi)+(sin(8pix))/(8pi) + C`

Explanation:

We seek:

`I = int \ 8cos^4 2pit \ dt`

We can utilise the double angle identity:

`cos 2A -= cos^2A - sin^2A`
`\ \ \ \ \ \ \ \ \ \ = cos^2A - (1-cos^2A)`
`\ \ \ \ \ \ \ \ \ \ = 2cos^2A - 1`

From which we get:

`cos^2A -= 1/2(1+cos2A)`

And squaring we get:

`cos^4A -= 1/4(1+cos2A)^2`
`\ \ \ \ \ \ \ \ \ \ = 1/4(1+2cos2A+cos^2 2A)`
`\ \ \ \ \ \ \ \ \ \ = 1/4(1+2cos2A+1/2(1+cos4A))`
`\ \ \ \ \ \ \ \ \ \ = 1/8(2+4cos2A+(1+cos4A))`
`\ \ \ \ \ \ \ \ \ \ = 1/8(3+4cos2A+cos4A)`

To make thing easier to read we can also perform a simple substitution:

Let `u =2pit => (du)/dt = 2pi`

So our integral becomes:

`I = 8 int \ (cos^4 u) \ 1/(2pi) \ du`
`\ \ = 8/(2pi) int \ (cos^4 u) du`
`\ \ = 8/(2pi) int \ (1/8(3+4cos2u+cos4u)) \ du`
`\ \ = 1/(2pi) int \ 3+4cos2u+cos4u \ du`
`\ \ = 1/(2pi) {3u + 4/2sin2u+1/4sin4u } + C`
`\ \ = 3u/(2pi) + (2sin2u)/(2pi)+(1/4sin4u)/(2pi) + C`

And restraining the substitution we have:

`I = 3(2pix)/(2pi) + (2sin(4pix))/(2pi)+(1/4sin(8pix))/(2pi) + C`
`\ \ = 3x + (sin(4pix))/(pi)+(sin(8pix))/(8pi) + C`