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...
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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