Question

How do you integrate $\smallint \tan 3x + \sec 3xdx$ ?

Answer 1

Hint: To solve this integral expression, first we should know that this integral can be solved by the u-substitution method. And then we can solve this question in two parts to make the steps to solve easier.

Complete step-by-step solution:
The given integral can be solved by using a u-substitution, i.e..
$\smallint \tan \,udu = - \ln |\cos u| + C$ and
$\smallint \sec \,udu = \ln |\sec u + \tan u| + C$
Now, back to the given expression- $\smallint \tan 3x + \sec 3xdx$ :
We can let, $u = 3x$
Differentiate the above assumed $u = 3x$ :
$
   \Rightarrow du = 3dx \\
   \Rightarrow \dfrac{1}{3}du = dx \\
$
Now, substitute the upper values in the main given integral expression:
$
  \therefore \smallint \tan 3x + \sec 3xdx = \dfrac{1}{3}\smallint \tan udu + \dfrac{1}{3}\smallint \sec udu \\
  \Rightarrow - \dfrac{1}{3}\ln |\cos 3x| + \dfrac{1}{3}|\sec 3x + \tan 3x| + C \\
    \Rightarrow \dfrac{1}{3}|\sec 3x + \tan 3x| - \dfrac{1}{3}\ln |\cos 3x| + C \\
$

Note: In calculus, as we use the u-substitution method in this question, that is the convenient method to solve the complex integration by assuming one of them part as $u$ . When deciding which part of our function to call $u$ , we will want to look for a piece of your function that you can see that piece’s derivative somewhere else in the function.