Integrate cos^2x

Nov 2010
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Not sure how to tackle this:
$$\int_0^\frac{\pi}{4}\frac{x}{\cos^2(x)}$$
 
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SlipEternal

MHF Helper
Nov 2010
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$\displaystyle \left. \int_0^{\tfrac{\pi}{4}} \dfrac{dx}{\cos^2 x} = \tan x \right]_0^{\tfrac{\pi}{4}} = 1$
 
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SlipEternal

MHF Helper
Nov 2010
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There's also an x in the nominator.
I see. Then you are missing a dx.

$u = x, dv = \sec^2 x dx$
$du = dx, v = \tan x$

$\displaystyle \left. \int_0^{\tfrac{\pi}{4}} \dfrac{xdx}{\cos^2 x} = x\tan x\right]_0^{\tfrac{\pi}{4}} - \int_0^{\tfrac{\pi}{4}} \tan x dx$

$u = \cos x$
$du = -\sin x dx$
$u(0) = 1, u\left(\dfrac{\pi}{4} \right) = \dfrac{\sqrt{2}}{2}$

$\displaystyle \dfrac{\pi}{4} - \left. \int_0^{\tfrac{\pi}{4}} \tan x dx = \dfrac{\pi}{4} + \int_1^{\tfrac{\sqrt{2}}{2}} \dfrac{du}{u} = \dfrac{\pi}{4} + \ln u \right]_1^{\tfrac{\sqrt{2}}{2}} = \dfrac{\pi}{4} -\dfrac{1}{2} \ln 2$
 
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