Trigonometric Integral: Finding the Antiderivative of cotx + tanx

[Mentallic]

Homework Statement

If $f(x)=cotx+tanx$ then find $\int{f(x)}dx$

The Attempt at a Solution

I was able to manipulate (not sure if in the right direction though) the function and resulted with $$f(x)=\frac{2}{sin(2x)}$$

I've also tried re-arranging things in a different way, but came up with nothing useful (I think). So then I'm stuck, any ideas?

 

[danago]

You could possibly use the fact that (sin x)' = cos x, along with the idea that (ln f(x))' = f'(x) / f(x).

Much easier to integrate term by term rather than combine the expressions into a single trig term.

 

[Mentallic]

I'm not getting any form of f'(x)/f(x) and substituting u=sinx and du/dx=cosx doesn't seem like it gave me much either...

 

[danago]

Look at the cot x term --

cot x = cos x / sin x, which is exactly of the form f'(x) / f(x), since (sin x)' = cos x. A very similar idea applies to the tan x term.

 

[HallsofIvy]

That wasn't what was suggested.

$$cot(x)+ tan(x)= \frac{cos(x)}{sin(x)}+ \frac{sin(x)}{cos(x)}$$

Integrate by letting u= sin(x) in the first fraction and v= cos(x) in the second.

 

[Mentallic]

Ahh thanks for all the help!

Sorry, I kept looking at my manipulated expression.