# Thread: [SOLVED] Integrating trig functions

1. ## [SOLVED] Integrating trig functions

$\displaystyle \int{\sqrt{sec^2x}}$

The square root is giving me some problems here. I know that the identity of $\displaystyle sec^2 x$ is $\displaystyle tan x$, but how does the sqrt play here?

2. Originally Posted by kl.twilleger
$\displaystyle \int{\sqrt{sec^2x}}$
the square root cancels the square. (technically you would have to use the absolute value, but it usually doesn't cause problems if we forget it, especially for indefinite integrals).

so you want $\displaystyle \int \sec x~dx$

now there's an interesting integral ...if you don't know it by heart, that is

The square root is giving me some problems here. I know that the identity of $\displaystyle sec^2 x$ is $\displaystyle tan x$, but how does the sqrt play here?
the identity is $\displaystyle \sec^2 x = 1 + \tan^2x$. but you don't need that here

3. just integrate

$\displaystyle \int \sec{x}~dx$

To integrate $\displaystyle \sec{x}$ multiply by $\displaystyle \frac{\sec{x} + \tan{x}}{\sec{x} + \tan{x}}$

4. DUH!!!

ok... so the integration is between 0 and pi/4 so it will be

$\displaystyle ln(sec\frac{\pi}{4} + tan\frac{\pi}{4}) - ln(sec 0 + tan 0)$ and then just plug and chug

5. Originally Posted by kl.twilleger
DUH!!!

ok... so the integration is between 0 and pi/4 so it will be

$\displaystyle ln(sec\frac{\pi}{4} + tan\frac{\pi}{4}) - ln(sec 0 + tan 0)$ and then just plug and chug
yes