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

$$\mathrm{Calcul}\mathrm{I}={\underset{0}{\int }}^{\frac{\pi }{2}}\frac{\ln \left(\mathrm{cost}\right)}{1+{\sin }^2\mathrm{t}}\mathrm{dt}$$

Answered by qaz last updated on 22/Sep/23

$$I={\int }_0^{\pi /2}\frac{ln\cos t}{2-\cos {}^{2}t}dt=-\frac{1}{2}{\int }_0^{\pi /2}\frac{ln\sec {}^{2}t}{2\sec {}^{2}t-1}\sec {}^{2}tdt$$$$=-\frac{1}{2}{\int }_0^{\pi /2}\frac{ln(1+\tan {}^{2}t)}{1+2\tan {}^{2}t}\sec {}^{2}tdt=-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{ln(1+x^2)}{1+2x^2}dx$$$$=-\frac{1}{2\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{ln(1+\frac{x^2}{2})}{1+x^2}dx=-\frac{1}{2\sqrt{2}}{\int }_0^{\mathrm{\infty }}\frac{ln\left(\frac{1}{2}\right)+ln(2+x^2))}{1+x^2}dx$$$$=\frac{\pi }{4\sqrt{2}}ln2-\frac{1}{2\sqrt{2}}{\int }_0^{2}dt{\int }_0^{\mathrm{\infty }}\frac{dx}{(1+x^2)(t+x^2)}$$$$=\frac{\pi }{4\sqrt{2}}ln2-\frac{\pi }{4\sqrt{2}}{\int }_0^2\frac{1}{t-1}(1-\frac{1}{\sqrt{t}})dt$$$$=\frac{\pi }{4\sqrt{2}}(ln2-2ln(1+\sqrt{2}))$$

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