Answer:
[tex]\int\ {f(x)} \, dx = 9e^x + 2ln(sec(2x) + tan(2x)) + C[/tex]
Step-by-step explanation:
think of the function having two parts, [tex]9e^x[/tex] and [tex]4sec(2x)[/tex]
and integrate them separately.
[tex]\int\ {9e^x} \, dx \\[/tex]
since 9 is a constant you
[tex]9\int\ {e^x} \, dx\\ [/tex]
[tex]9e^x[/tex]
[tex]\int\ {4sec(2x)} \, dx \\4\int\ {sec(2x)} \, dx \\[/tex]
we can use u-substitution [tex]u = 2x[/tex] and [tex]du =2dx[/tex]
[tex]4\int\ {sec(u)} \, \frac{du}{2} \\2\int\ {sec(u)} \, du\\[/tex]
think of it as only integrating sec(x)
[tex]2(ln(sec(u) + tan(u)))\\2(ln(sec(2x) + tan(2x)))\\[/tex]
Answer:: [tex]9e^x + 2ln(sec(2x) + tan(2x)) + C\\[/tex]
