Brain Fade About Trig w/Differential

[misskitty]

Multiply 1/(square root of 3) by the (square root of 3)/(square root of 3). Leaving you with the square root of 3 in the numerator and 3 in the denominator. Then multiply that by 2/1 (same thing as 2) to give you 2(square root of 3)/3.

How did you think I rationalized?

 

[Hootenanny]

Why have you multiplied by 2/1? That has changed the fraction. Once you have mulitplied by $\frac{\sqrt{3}}{\sqrt{3}}$ the fraction is now rational. When you mulitplied by $\frac{2}{1}$ you changed the fraction!

 

[VietDao29]

Why have you multiplied by 2/1? That has changed the fraction. Once you have mulitplied by $\frac{\sqrt{3}}{\sqrt{3}}$ the fraction is now rational. When you mulitplied by $\frac{2}{1}$ you changed the fraction!

?
What? What do you mean? I just don't get this!
$$\mathop{\int} \limits_{-\frac{\pi}{6}} ^ {\frac{\pi}{6}} \sec ^ 2 x dx = \tan \left( \frac{\pi}{6} \right) - \tan \left( - \frac{\pi}{6} \right) = 2 \tan \left( \frac{\pi}{6} \right) = \frac{2}{\sqrt{3}} = \frac{2 \sqrt{3}}{3}$$
What misskitty did is, needless to say, completely correct!

 

[Hootenanny]

I apologise, it was just in her post she said she mulitplied by two, I didn't understand where it came from. I only saw the $\frac{1}{\sqrt{3}}$, not the $\frac{2}{\sqrt{3}}$