# `int sin(-7x)cos(6x) dx` Find the indefinite integral

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 sin(-7x)cos(6x) dx` or `intcos(6x)sin(-7x) dx`   has a integrand in a form of trigonometric function. To evaluate this, we apply the identity:

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

The integral becomes:

`intcos(6x)sin(-7x) dx = int[sin(6x+(-7x)) -sin(6x-(-7x))]/2dx`

`= int[sin(6x-7x) -sin(6x+7x)]/2dx`

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

`int[sin(6x-7x) -sin(6x+7x)]/2dx= 1/2int[sin(6x-7x) -sin(6x+7x)]dx`

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

`1/2 *[int (sin(6x-7x))dx - int sin(6x+7x)dx]`

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

For the integral: `intsin(6x-7x)dx` , we let `u = 6x-7x=-x` then `du= - dx` or `(-1)du =dx` .

`intsin(6x-7x)dx=intsin(-x) dx`

`=intsin(u) *(-1)du`

`=(-1) int sin(u)du`

`=(-1)(-cos(u) )+C`

`=cos(u) +C`

Plug-in `u =-x` on `cos(u) +C` , we get:

`intsin(6x-7x)dx= cos(-x) +C`

For the integral: `intsin(6x+7x)dx` , we let `u = 6x+7x=13x` then `du= 13 dx` or `(du)/13 =dx` .

`intsin(6x+7x)dx=intsin(13x) dx`

` =intsin(u) *(du)/13`

` = 1/13 int sin(u)du`

`= 1/13( -cos(u))+C or -1/13cos(u) +C`

Plug-in `u =13x` on `-1/13 cos(u) +C` , we get:

`intsin(6x+7x)dx= -1/13 cos(13x) +C`

Combing the results, we get the indefinite integral as:

`intsin(-7x)cos(6x) dx= 1/2*[ cos(-x) -(-1/13 cos(13x))] +C`

or   `1/2 cos(-x) +1/26 cos(13x) +C`

Since cosine is an even function,  `cos(-x) = cos(x)` , so we get:

`intsin(-7x) cos(6x)dx=1/2 cos(x) +1/26 cos(13x) +C`

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