# Integrate 1/9+cosx dx?

Jun 20, 2018

$\frac{1}{9} x + \sin \left(x\right) + c$

#### Explanation:

The integral is linear, i.e.

$\setminus \int a f \left(x\right) + b g \left(x\right) = a \setminus \int f \left(x\right) + b \setminus \int g \left(x\right)$

so, in particular,

$\setminus \int f \left(x\right) + g \left(x\right) = \setminus \int f \left(x\right) + \setminus \int g \left(x\right)$

In this case,

$\setminus \int \frac{1}{9} + \cos \left(x\right) = \setminus \int \frac{1}{9} + \setminus \int \cos \left(x\right)$

The integral of every costant $k$ is

$\setminus \int k = k x + c$

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

$\setminus \int \cos \left(x\right) = \sin \left(x\right)$

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.

$\setminus \int \frac{1}{9} + \cos \left(x\right) = \setminus \int \frac{1}{9} + \setminus \int \cos \left(x\right) = \frac{1}{9} x + \sin \left(x\right) + c$