Integrating by eliminating square root

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Oct 6th 2009, 03:28 PM
genlovesmusic09

Integrating by eliminating square root

$\int_{\frac{{ - \pi }}<br /> {4}}^{\frac{\pi }<br /> {4}} {\sqrt {1 + {{\tan }^2}x} dx}<br />$

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

Quote:

Originally Posted by genlovesmusic09

$\int_{\frac{{ - \pi }}<br /> {4}}^{\frac{\pi }<br /> {4}} {\sqrt {1 + {{\tan }^2}x} dx}<br />$

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

$\int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}}<br /> {{{{\cos }^2}x}}} } } <br />$

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

Quote:

Originally Posted by genlovesmusic09

$\int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}}<br /> {{{{\cos }^2}x}}} } } <br />$

i don't see the next step

$\sqrt{(something)^2} = |something|<br />$
Oct 6th 2009, 04:00 PM
genlovesmusic09

wow... i am overthinking
thanks

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