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

Quote:

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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

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

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

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

Can you take it from here?

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

$\displaystyle \int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}}
{{{{\cos }^2}x}}} } }
$

i don't see the next step

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

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$\displaystyle \sqrt{(something)^2} = |something|
$

wow... i am overthinking
thanks