pi-4-pi-4-sec-2-x-tgx-2-dx-

Question Number 141387 by cesarL last updated on 18/May/21

\int_{- π / 4}^{π / 4} {({s e c}^{2} x + t g x)}^{2} d x

Answered by MJS_new last updated on 18/May/21

\underset{- π / 4}{\int^{π / 4}} {(\sec^{2} x + \tan x)}^{2} d x =

[t = \tan x \to d x = \cos^{2} x d t]

= \underset{- 1}{\int^{1}} \frac{{(t^{2} + t + 1)}^{2}}{t^{2} + 1} d t = \underset{- 1}{\int^{1}} (t^{2} + 2 t + 2 - \frac{1}{t^{2} + 1}) d t =

= {[\frac{t^{3}}{3} + t^{2} + 2 t - \arctan t]}_{- 1}^{1} = \frac{14}{3} - \frac{π}{2}

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