$\displaystyle \int_{\frac{{ - \pi }}

{4}}^{\frac{\pi }

{4}} {\sqrt {1 + {{\tan }^2}x} dx}

$

I'm not sure where to begin

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- Oct 6th 2009, 03:28 PMgenlovesmusic09Integrating 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 PMartvandalay11
- Oct 6th 2009, 03:39 PMpickslides
$\displaystyle 1 + {{\tan }^2}x = \sec^2x$

- Oct 6th 2009, 03:49 PMgenlovesmusic09
$\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 PMskeeter
- Oct 6th 2009, 04:00 PMgenlovesmusic09
wow... i am overthinking

thanks