$\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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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?
$\displaystyle 1 + {{\tan }^2}x = \sec^2x$
$\displaystyle \int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}} {{{{\cos }^2}x}}} } } $ i don't see the next step
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| $
wow... i am overthinking thanks
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