How do you integrate $int e^(sec2x)sec2xtan2xdx$ from $[pi/3,pi/2]$ ?

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Answer 1 · 2017-04-21T05:36:28

$1/2 * (e^(-1) - e^(-2))$

$d/dx(sec 2x) = d/dx (1/(cos 2x)) = - 1/(cos 2x)^2 * (-sin 2x) * 2$
$= sec 2x * tan 2x * 2$

Hence

$d/dx e^(sec 2x) = e^(sec 2x) * d/dx(sec 2x) = e^(sec 2x) * sec 2x * tan 2x * 2$

So

$int_(pi/3)^(pi/2) e^(sec2x)sec2xtan2xdx = 1/2 e^(sec 2x)|_(pi/3)^(pi/2) = 1/2 * (e^(-1) - e^(-2))$

How do you integrate int e^(sec2x)sec2xtan2xdxint e^(sec2x)sec2xtan2xdx from [pi/3,pi/2][pi/3,pi/2]?