$\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 \int_{\frac{{ - \pi }}
{4}}^{\frac{\pi }
{4}} {\sqrt {1 + {{\tan }^2}x} dx}
$
I'm not sure where to begin
$\displaystyle 1 + {{\tan }^2}x = \sec^2x$
$\displaystyle \int {\sqrt {{{\sec }^2}x} dx = \int {\sqrt {\frac{{dx}}
{{{{\cos }^2}x}}} } }
$
i don't see the next step
wow... i am overthinking
thanks