Calcul-I-pi-2-0-ln-cost-1-sin-2-t-dt-

Question Number 197550 by Erico last updated on 21/Sep/23

Calcul I = {\int_{0}}^{\frac{π}{2}} \frac{\ln (cost)}{1 + \sin^{2} t} dt

Answered by qaz last updated on 22/Sep/23

I = \int_{0}^{π / 2} \frac{l n \cos t}{2 - \cos^{2} t} d t = - \frac{1}{2} \int_{0}^{π / 2} \frac{l n \sec^{2} t}{2 \sec^{2} t - 1} \sec^{2} t d t

= - \frac{1}{2} \int_{0}^{π / 2} \frac{l n (1 + \tan^{2} t)}{1 + 2 \tan^{2} t} \sec^{2} t d t = - \frac{1}{2} \int_{0}^{\infty} \frac{l n (1 + x^{2})}{1 + 2 x^{2}} d x

= - \frac{1}{2 \sqrt{2}} \int_{0}^{\infty} \frac{l n (1 + \frac{x^{2}}{2})}{1 + x^{2}} d x = - \frac{1}{2 \sqrt{2}} \int_{0}^{\infty} \frac{l n (\frac{1}{2}) + l n (2 + x^{2}))}{1 + x^{2}} d x

= \frac{π}{4 \sqrt{2}} l n 2 - \frac{1}{2 \sqrt{2}} \int_{0}^{2} d t \int_{0}^{\infty} \frac{d x}{(1 + x^{2}) (t + x^{2})}

= \frac{π}{4 \sqrt{2}} l n 2 - \frac{π}{4 \sqrt{2}} \int_{0}^{2} \frac{1}{t - 1} (1 - \frac{1}{\sqrt{t}}) d t

= \frac{π}{4 \sqrt{2}} (l n 2 - 2 l n (1 + \sqrt{2}))

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