Question

Integrate 1/9+cosx dx?

Answer 1

`1/9x + sin(x) + c`

Explanation:

The integral is linear, i.e.

`\int af(x)+bg(x) = a \int f(x) + b \int g(x)`

so, in particular,

`\int f(x)+g(x) = \int f(x) + \int g(x)`

In this case,

`\int 1/9+cos(x) = \int 1/9 + \int cos(x)`

The integral of every costant `k` is

`\int k = kx+c`

while the integral of the cosine function is the sine function:

`\int cos(x) = sin(x)`

both results can easily be shown by deriving: if you derive a first degree polynomial you get a constant, while if you derive the sine function, you get the cosine function.

So, the final answer is

`\int 1/9+cos(x) = \int 1/9 + \int cos(x) = 1/9x + sin(x) + c`