# Integrating by eliminating square root

• Oct 6th 2009, 03:28 PM
genlovesmusic09
Integrating by eliminating square root
$\displaystyle \int_{\frac{{ - \pi }} {4}}^{\frac{\pi } {4}} {\sqrt {1 + {{\tan }^2}x} dx}$

I'm not sure where to begin
• Oct 6th 2009, 03:35 PM
artvandalay11
Quote:

Originally Posted by genlovesmusic09
$\displaystyle \int_{\frac{{ - \pi }} {4}}^{\frac{\pi } {4}} {\sqrt {1 + {{\tan }^2}x} dx}$

I'm not sure where to begin

$\displaystyle sin^2x+cos^2x=1$

Let's divide by $\displaystyle cos^2x$

$\displaystyle tan^2x+1=sec^2x$

Can you take it from here?
• Oct 6th 2009, 03:39 PM
pickslides
$\displaystyle 1 + {{\tan }^2}x = \sec^2x$
• Oct 6th 2009, 03:49 PM
genlovesmusic09
$\displaystyle \int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}} {{{{\cos }^2}x}}} } }$

i don't see the next step
• Oct 6th 2009, 03:56 PM
skeeter
Quote:

Originally Posted by genlovesmusic09
$\displaystyle \int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}} {{{{\cos }^2}x}}} } }$

i don't see the next step

$\displaystyle \sqrt{(something)^2} = |something|$
• Oct 6th 2009, 04:00 PM
genlovesmusic09
wow... i am overthinking
thanks