0-pi-2-1-sec-2-t-sec-t-1-sec-t-2-2-dt-

Question Number 136969 by liberty last updated on 28/Mar/21

\int_{0}^{π / 2} \frac{(1 + \sec^{2} t) \sqrt{\sec t}}{{(1 + \sec t)}^{2} - 2} dt = ?

Answered by EDWIN88 last updated on 28/Mar/21

E = \int_{0}^{π / 2} \frac{(1 + \sec^{2} t) \sqrt{\sec t}}{{(1 + \sec t)}^{2} - 2} dt

E = \int_{0}^{π / 2} \frac{1 + \cos^{2} t}{\sqrt{\cos t} (1 + 2 \cos t - \cos^{2} t)} dt

making change of variable u = \sqrt{\cos t}

E = \int_{1}^{0} \frac{u^{4} + 1}{u (1 + {2 u}^{2} - u^{4})} . (- \frac{2 u}{\sqrt{1 - u^{4}}}) du

E = 2 \int_{0}^{1} \frac{u^{4} + 1}{(1 + {2 u}^{2} - u^{4}) \sqrt{1 - u^{4}}} du

E = 2 \int_{0}^{1} \frac{1 + u^{- 4}}{(u^{- 2} + 2 - u^{2}) \sqrt{u^{- 2} - u^{2}}} du

set \sqrt{u^{- 2} - u^{2}} = v

E = 2 \int_{0}^{\infty} \frac{dv}{v^{2} + 2} = \sqrt{2} {[\arctan (\frac{v}{\sqrt{2}})]}_{0}^{\infty}

E = \frac{π}{\sqrt{2}} .

Commented by liberty last updated on 28/Mar/21

great

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