We can start by rewriting our integral as
int (1/cosx)+color(blue)((sinx/cosx)) dx∫(1cosx)+(sinxcosx)dx
What I have in blue is equal to tanxtanx, so now we have the following integral:
int (1/cosx+tanx) dx∫(1cosx+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)∫(1cosx)dx+∫(tanx)dx
Integrating 1/cosx1cosx
1/cosx1cosx is equivalent to secxsecx. Now, we have the integral
color(springgreen)(int (secx) dx)∫(secx)dx, which evaluates to
color(springgreen)(ln|tanx+secx| +C)ln|tanx+secx|+C
Remember, this is one part of our integral.
Integrating tanxtanx
We can rewrite tanxtanx as sinx/cosxsinxcosx. Doing this, we get
color(orchid)(int (sinx/cosx) dx)∫(sinxcosx)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)u=cosx
From this, we know that
color(orchid)(du=-sinx)du=−sinx
Since we have sinxsinx, not -sinx−sinx, we can put a negative in front of the integral. Plugging in, we get
color(orchid)(-int (1/u du))−∫(1udu)
which evaluates to
color(orchid)(-ln|u|)−ln|u|
Now, we need to back-substitute (plug uu back in). We get
color(orchid)(-ln|cosx|+C)−ln|cosx|+C
Thus, the integral of ((1+sinx)/cosx) dx(1+sinxcosx)dx is
bar (ul( |color(white)(2/2)(ln|tanx+secx|-ln|cosx|+C)color(white)(2/2) |
Hope this helps!
