Integrate 1/9+cosx dx?

1 Answer
KillerBunny
Jun 20, 2018

#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#

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