Compute the area of the region bounded by the curves $y = t a n x$ and $y = t a n^{2} (\frac{- π}{2} < x < \frac{π}{2}) .$

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Solution

$y = tan x a n d y = {tan}^{2} x$

$A r e a = 2 {\int_{0}^{\frac{π}{4}} tan x d x + \int_{\frac{π}{4}}^{\frac{π}{2}} ({sec}^{2} x - 1) d x}$

$= 2 {| log sec x |}_{0}^{\frac{π}{4}} + 2 {| tan x - x |}_{\frac{π}{4}}^{\frac{π}{2}}$

$= 2 log (\sqrt{2}) + 2 (- \frac{π}{2} - 1 + \frac{π}{4})$

$= log 2 + 2 (- \frac{π}{2} - 1)$

$= log 2 - 2 (1 + \frac{π}{2}) s q . u n i t$

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