`int cos(2x)cos(6x) dx` Find the indefinite integral

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Indefinite integrals are written in the form of `int f(x) dx = F(x) +C`

 where: `f(x)` as the integrand

          `F(x)` as the anti-derivative function 

           `C`  as the arbitrary constant known as constant of integration

For the given problem: `int cos(6x)cos(2x) dx` has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:

`cos(A)cos(B) =[cos(A+B) +cos(A-B)]/2`

The integral becomes:

`int cos(6x)cos(2x) dx = int[cos(6x+2x) +cos(6x-2x)]/2dx`

 Apply the basic properties of integration: `int c*f(x) dx= c int f(x) dx` .

`int[cos(6x+2x) +cos(6x-2x)]/2dx = 1/2int[cos(6x+2x) +cos(6x-2x)]dx`

 Apply the basic integration property: `int (u+v) dx = int (u) dx + int (v) dx` .

`1/2 *[intcos(6x+2x) dx+int cos(6x-2x)dx]`

Then apply u-substitution to be able to apply integration formula for cosine function: `int cos(u) du= sin(u) +C` .

For the integral: int cos(6x+2x) dx, we let `u = 6x+2x =8x` then `du= 8 dx` or `(du)/8 =dx` .

`intcos(6x+2x) dx=intcos(8x) dx`

                                 `=intcos(u) *(du)/8`

                                `= 1/8 int cos(u)du`

                                 `= 1/8 sin(u) +C`

Plug-in `u =8x` on `1/8 sin(u) +C` , we get:

`intcos(6x+2x) dx=1/8 sin(8x) +C`

 For the integral: `intcos(6x-2x) dx` , we let `u = 6x-2x =4x` then `du= 4 dx` or `(du)/4 =dx` .

`intcos(6x-2x) dx=intcos(4x) dx`

                                `=intcos(u) *(du)/4`

                                `= 1/4 int cos(u)du`

                                `= 1/4 sin(u) +C`

Plug-in `u =4x` on `1/4 sin(u) +C` , we get:

`intcos(6x-2x) dx=1/4 sin(4x) +C`

Combing the results , we get the indefinite integral as:

`1/2 *[intcos(6x+2x) dx+int cos(6x-2x)dx] = 1/2*[1/8 sin(8x) +1/4 sin(4x)] +C`

or ` 1/16 sin(8x) +1/8 sin(4x) +C`

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