We can start by rewriting our integral as
#int (1/cosx)+color(blue)((sinx/cosx)) dx#
What I have in blue is equal to #tanx#, so now we have the following integral:
#int (1/cosx+tanx) dx#
The integral of the sum is equal to the sum of the integrals, so we can rewrite the above expression as
#color(springgreen)(int (1/cosx)dx)+color(orchid)(int (tanx)dx)#
Integrating #1/cosx#
#1/cosx# is equivalent to #secx#. Now, we have the integral
#color(springgreen)(int (secx) dx)#, which evaluates to
#color(springgreen)(ln|tanx+secx| +C)#
Remember, this is one part of our integral.
Integrating #tanx#
We can rewrite #tanx# as #sinx/cosx#. Doing this, we get
#color(orchid)(int (sinx/cosx) dx)#
Whenever we see a function and its derivative, it's a good idea to use u-substitution. Let's define
#color(orchid)(u=cosx)#
From this, we know that
#color(orchid)(du=-sinx)#
Since we have #sinx#, not #-sinx#, we can put a negative in front of the integral. Plugging in, we get
#color(orchid)(-int (1/u du))#
which evaluates to
#color(orchid)(-ln|u|)#
Now, we need to back-substitute (plug #u# back in). We get
#color(orchid)(-ln|cosx|+C)#
Thus, the integral of #((1+sinx)/cosx) dx# is
#bar (ul( |color(white)(2/2)(ln|tanx+secx|-ln|cosx|+C)color(white)(2/2) |#
Hope this helps!
