# Integrating by eliminating square root

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

I'm not sure where to begin

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

Let's divide by $cos^2x$

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

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

i don't see the next step

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