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Question Number 197550 by Erico last updated on 21/Sep/23
Calcul I=∫^( (π/2)) _( 0) ((ln(cost))/(1+sin^2 t))dt
$$\mathrm{Calcul}\:\:\:\mathrm{I}=\underset{\:\mathrm{0}} {\int}^{\:\frac{\pi}{\mathrm{2}}} \frac{\mathrm{ln}\left(\mathrm{cost}\right)}{\mathrm{1}+\mathrm{sin}^{\mathrm{2}} \mathrm{t}}\mathrm{dt} \\ $$
Answered by qaz last updated on 22/Sep/23
I=∫_0 ^(π/2) ((lncos t)/(2−cos^2 t))dt=−(1/2)∫_0 ^(π/2) ((lnsec^2 t)/(2sec^2 t−1))sec^2 tdt =−(1/2)∫_0 ^(π/2) ((ln(1+tan^2 t))/(1+2tan^2 t))sec^2 tdt=−(1/2)∫_0 ^∞ ((ln(1+x^2 ))/(1+2x^2 ))dx =−(1/(2(√2)))∫_0 ^∞ ((ln(1+(x^2 /2)))/(1+x^2 ))dx=−(1/(2(√2)))∫_0 ^∞ ((ln((1/2))+ln(2+x^2 )))/(1+x^2 ))dx =(π/(4(√2)))ln2−(1/(2(√2)))∫_0 ^2 dt∫_0 ^∞ (dx/((1+x^2 )(t+x^2 ))) =(π/(4(√2)))ln2−(π/( 4(√2)))∫_0 ^2 (1/(t−1))(1−(1/( (√t))))dt =(π/(4(√2)))(ln2−2ln(1+(√2)))
$${I}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\mathrm{cos}\:{t}}{\mathrm{2}−\mathrm{cos}\:^{\mathrm{2}} {t}}{dt}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\mathrm{sec}\:^{\mathrm{2}} {t}}{\mathrm{2sec}\:^{\mathrm{2}} {t}−\mathrm{1}}\mathrm{sec}\:^{\mathrm{2}} {tdt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{ln}\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {t}\right)}{\mathrm{1}+\mathrm{2tan}\:^{\mathrm{2}} {t}}\mathrm{sec}\:^{\mathrm{2}} {tdt}=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\mathrm{1}+\mathrm{2}{x}^{\mathrm{2}} }{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{{ln}\left(\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}=−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\infty} \frac{\left.{ln}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)+{ln}\left(\mathrm{2}+{x}^{\mathrm{2}} \right)\right)}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\mathrm{2}−\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{2}} {dt}\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left({t}+{x}^{\mathrm{2}} \right)} \\ $$$$=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\mathrm{2}−\frac{\pi}{\:\mathrm{4}\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{1}}{{t}−\mathrm{1}}\left(\mathrm{1}−\frac{\mathrm{1}}{\:\sqrt{{t}}}\right){dt} \\ $$$$=\frac{\pi}{\mathrm{4}\sqrt{\mathrm{2}}}\left({ln}\mathrm{2}−\mathrm{2}{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\right) \\ $$

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