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Math Help - evaluate the integral

  1. #1
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    evaluate the integral

    upper limit Pi/4

    lower limit 0

    (2tanx)/((cosx)^2) dx
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  2. #2
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    2\int \frac{\tan x}{\cos^2 x}=2\int\frac{\sin x}{\cos^3 x}\;dx

    Let u=\cos x

    -2\int\frac{du}{u^3}=\frac{1}{\cos^2x}+C=\sec^2{x}+  C

    Thus, \int_0^\frac{\pi}{4} \frac{2\tan x}{\cos^2x}\;dx=\sec^2\left(\frac{\pi}{4}\right)-\sec^2(0)=(\sqrt{2})^2-(1)^2=1
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