# Thread: Need help with Trig Integral

1. ## Need help with Trig Integral

Hello,

I've been trying to work the following problem, but I am getting lost from the start. I'm pretty sure if I could get it started I wouldn't have a problem with it.

$\displaystyle (1+4cot(x))/(4-cot(x))$

2. $\displaystyle \frac{1 + 4\cot(x)}{4 - \cot(x) }$

$\displaystyle = \frac{ \tan(x) \cdot (1 + 4\cot(x) )}{ \tan(x) \cdot ( 4 - \cot(x))}$

$\displaystyle = \frac{ \tan(x) + 4 }{ 4\tan(x) - 1 }$

$\displaystyle = -\frac{ \tan(x) + 4}{ 1 - 4\tan(x) }$

then use the formula

$\displaystyle \tan(A + B) = \frac{ \tan(A) + \tan(B) }{ 1 - \tan(A)\tan(B) }$

By substituting $\displaystyle A = x , B = \tan^{-1}(4)$

We obtain $\displaystyle \frac{1 + 4\cot(x)}{4 - \cot(x) } = - \tan( x + \tan^{-1}(4) )$

Wow , it is actually a tagent function so the integral is log. function :

$\displaystyle \ln{ \left [ \cos( x + \tan^{-1}(4) ) \right ]} + C$

3. Thank you!

Multiplying by 1 is something I would've never seen in my life. I always seem to forget that as an option.