Find the integral of \[\cos (3x)\] ?

Hint: Integral of trigonometric functions are already defined but for basic identity, if you find any changes in the angle given the you have to first expand or convert the given identity into basic then you can further integrate the function, for example integral of “sinx” is “-cosx+c”, “c” is used as integral constant.

Complete step-by-step answer:
The given question is \[\cos (3x)\]
After implying integral sign we get:
 \[ = \int {\cos \left( {3x} \right)dx} \]
Here we have to use substitution method to solve the question, in substitution method we have to assume a variable say “u” such that “3x” is equal to the variable to “u”
Now on further solving we get,
 \[
  \dfrac{{du}}{{dx}} = 3\dfrac{{d(x)}}{{dx}} \\
  \dfrac{{du}}{{dx}} = 3(1) \\
  dx = \dfrac{1}{3}du \;
\]
Now substituting these values we get:
 \[ = \int {\dfrac{1}{3}\cos (u)du} \]
Now we have to solve this integral with respect to “u”
We get;
 \[
   = \int {\dfrac{1}{3}\cos (u)du} \\
   = \dfrac{1}{3}\sin u\,(\int {\cos dx = \sin x} ) \\
   = \dfrac{1}{3}\sin (3x) \;
 \]
This is our required integral.
So, the correct answer is “$\dfrac{1}{3}\sin (3x)$”.

Note: This is the easiest way to solve this question, but if you find any difficulty then you can try another method that is, convert the equation in term of “sinx” then solve for that and the same result you will be getting after you again change the “sinx” term in “cosx” term. Changing it into “sinx” is not a good option here because the expansion of “cos3x” is already simple to solve and get the result.