# Integrate sec x

• Jan 6th 2010, 08:42 AM
roshanhero
Integrate sec x
How can we integrate sec x? Please give me only hint.
• Jan 6th 2010, 08:44 AM
Drexel28
Quote:

Originally Posted by roshanhero
How can we integrate sec x? Please give me only hint.

Let $x=2\arctan(z)$. That is the most fundamental and fool-proof way. There are cleverer ones.
• Jan 6th 2010, 09:09 AM
Calculus26
another way is to multiply and divide sec(x) by (secx +tan(x) to obtain

[sec^2(x) +sec(x)tan(x)]/[sec(x)+tan(x)]

Then let u = sec(x) +tan(x)

du = [ sec^2(x) +sec(x)tan(x)]dx

you end up with integral[du/u] = ln(u)
• Jan 6th 2010, 09:10 AM
Krizalid
Multiply top and bottom by $\sec x+\tan x.$ (Got beaten.)
• Jan 6th 2010, 09:42 AM
Shanks
multiply top and bottom by $\cos x$.
• Jan 6th 2010, 09:46 AM
Calculus26
Don't really see what multiplying top and bottom by cos(x) does for you??
• Jan 6th 2010, 09:55 AM
Shanks
Quote:

Originally Posted by Calculus26
Don't really see what multiplying top and bottom by cos(x) does for you??

After that, substitute $\cos ^2x$ with $1-\sin ^2x$, then seperate the fraction into two parts, I think you can take it from here now.
• Jan 6th 2010, 09:56 AM
Drexel28
Quote:

Originally Posted by Calculus26
Don't really see what multiplying top and bottom by cos(x) does for you??

$\int\frac{dx}{\cos(x)}=\int\frac{\cos(x)}{\cos^2(x )}dx=\int\frac{\cos(x)}{1-\sin^2(x)}dx$

Let $\sin(x)=z\implies dz=\cos(x)\text{ }dx$

So our integral becomes $\int\frac{dz}{1-z^2}$