# integrals of secx and 1/cosx

• Dec 7th 2008, 10:01 AM
hmmmm
integrals of secx and 1/cosx
why are the integrals of secx and 1/cosx not the same??
• Dec 7th 2008, 10:04 AM
galactus
They are the same. Because sec(x)=1/cos(x)

What difference do you see?. Please explain so I may know what you're getting at.
• Dec 7th 2008, 10:11 AM
hmmmm
• Dec 7th 2008, 10:15 AM
galactus
They are equivalent, just in a different form.

$\displaystyle tan(\frac{x}{2}+\frac{\pi}{4})=sec(x)+tan(x)$

Can you prove it?.
• Dec 7th 2008, 10:24 AM
Moo
Quote:

Originally Posted by hmmmm

And what if $\displaystyle \tan \left(\frac{ax}{2}+\frac \pi 4\right)=\sec(ax)+\tan(ax)$ ? (Tongueout)

$\displaystyle \tan \left(\frac{ax}{2}+\frac \pi 4\right)=\frac{\tan \left(\frac{ax}{2}\right)+\tan \left(\frac \pi 4\right)}{1-\tan \left(\frac{ax}{2}\right)\tan \left(\frac \pi 4\right)}$
We know that $\displaystyle \tan \frac \pi 4=1$. hence :

$\displaystyle \tan \left(\frac{ax}{2}+\frac \pi 4\right)=\frac{1+\tan \left(\frac{ax}{2}\right)}{1-\tan \left(\frac{ax}{2}\right)}$

Multiply by $\displaystyle \frac{1+\tan \left(\frac{ax}{2}\right)}{1+\tan \left(\frac{ax}{2}\right)}$ :

$\displaystyle =\frac{\left(1+\tan \left(\frac{ax}{2}\right)\right)^2}{1-\tan^2 \left(\frac{ax}{2}\right)}$

Now substitute $\displaystyle \tan^2t=\frac{1}{\cos^2t}-1$

Then multiply both numerator and denominator by cosē and use this formula : $\displaystyle 2 \cos^2t-1=\cos(2t)$