# Thread: more trigonomentry

1. ## more trigonomentry

$$\frac{tan^3(x)+tan(x)}{tan^3(x)+3tan^2(x)+2tan(x )+6}dx$$

Help section in the book tells to use substitution t=tan(x) so $$dx=cos^2(t) dt$$:
$$\frac{t^3+t}{t^3+3t^2+2t+6}cos^2(t) dt$$

How do i continue from here? Is this a good path to take?

2. ## Re: more trigonomentry

$$d x =\cos ^2x d t$$

3. ## Re: more trigonomentry

Ah yes, i have trouble with those.

I can't use $$cos^2(x)$$ right? Coz it's x and im trying to use t. How do i replace dx then?

4. ## Re: more trigonomentry

Originally Posted by TriForce
Ah yes, i have trouble with those.

I can't use $$cos^2(x)$$ right? Coz it's x and im trying to use t. How do i replace dx then?
$dx = \cos^2 x dt$

If $t = \tan x = \dfrac{\text{opp}}{\text{adj}}$, then $\cos x = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{1}{\sqrt{t^2+1}}$.

So, $dx = \dfrac{dt}{1+t^2}$

5. ## Re: more trigonomentry

Solved it, thanks