# Find the indefinite integral of `cos^2(2x+4)` ?

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You need to come up with the following substitution, such that:

`2x + 4 = y => (2x + 4)' = y' => 2dx = dy => dx = (dy)/2`

Changing the variable yields:

`int cos^2(2x + 4) dx = int cos^2 y *(dy)/2`

You should replace `1 - sin^2 y` for `cos^2 y` , such that:

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

Using the property of linearity yields:

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

You need to use half angle identity, such that:

`sin^2 y = (1 - cos 2y)/2`

`int (1 - sin^2 y) (dy)/2 = (1/2)*y - int (1 - cos 2y)/4 dy`

Splitting the integral` int (1 - cos 2y)/4 dy ` in two simpler integrals yields:

`int (1 - sin^2 y)/2 dy = (1/2)*y - int (1/4) dy + int (cos 2y)/4 dy`

`int (1 - sin^2 y)/2 dy = (1/2)*y - (1/4)*y + (sin 2y)/8 + c`

Replacing back `2x + 4 ` for y yields:

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

**Hence, evaluating the given indefinite integral, yields **`int cos^2(2x + 4)dx = x/2 + 1 + (sin(4x + 8))/8 + c`

The integral `int cos^2(2x+4) dx` has to be determined.

`int cos^2(2x+4) dx `

use the formula `cos 2x = 2cos^2x - 1 => cos^2x = (1 + cos 2x)/2 `

=> `int (1 + cos (4x + 8)/2 dx `

=> `(1/2) int 1 dx + (1/2) int cos (4x + 8) dx `

=> `x/2 + (1/2) int cos (4x + 8) dx `

let `4x + 8 = y => dy = 4 dx `

=> `x/2 + (1/2) int (1/4)*cos y dy `

=> `x/2 + sin y/8 `

=> `x/2 + sin (4x + 8)/8 `

**The integral `int cos^2(2x+4) dx = x/2 + sin (4x + 8)/8` **