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Question Number 208280 by Shrodinger last updated on 10/Jun/24

L=∫_0 ^(4/π) ln(cosx)dx

$${L}=\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\pi}} {ln}\left({cosx}\right){dx} \\ $$

Answered by Berbere last updated on 10/Jun/24

=∫_0 ^(π/4) ((ln(((cos(x))/(sin(x))).sin(x)cos(x))dx)/2)=(1/2)∫_0 ^(π/4) ln(cot(x))dx+(1/2)∫_0 ^(π/4) ln(((sin(2x))/2))dx tan(x)→t;2x→t =−(1/2)∫_0 ^1 ((ln(t))/(1+t^2 ))dt+(1/4)∫_0 ^(π/2) ln(sin(y))dy−((ln(2))/2).(π/4) =−(1/2)∫_0 ^1 Σ_(n≥0) (−1)^n ln(t)t^(2n) dt+(1/4).−(π/2)ln(2)−((πln(2))/8) =−(1/2)Σ_(n≥0) −(1/((2n+1)^2 ))−((πln(2))/4);G=(((−1)^n )/((2n+1)^2 ))=β(−1) Catalane constant =(1/2)G−((πln(2))/4)

$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{ln}\left(\frac{{cos}\left({x}\right)}{{sin}\left({x}\right)}.{sin}\left({x}\right){cos}\left({x}\right)\right){dx}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left({cot}\left({x}\right)\right){dx}+\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {ln}\left(\frac{{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}\right){dx} \\ $$$${tan}\left({x}\right)\rightarrow{t};\mathrm{2}{x}\rightarrow{t} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{ln}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}+\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({y}\right)\right){dy}−\frac{{ln}\left(\mathrm{2}\right)}{\mathrm{2}}.\frac{\pi}{\mathrm{4}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{n}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{{n}} {ln}\left({t}\right){t}^{\mathrm{2}{n}} {dt}+\frac{\mathrm{1}}{\mathrm{4}}.−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{8}} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\underset{{n}\geqslant\mathrm{0}} {\sum}−\frac{\mathrm{1}}{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{4}};{G}=\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)^{\mathrm{2}} }=\beta\left(−\mathrm{1}\right)\:{Catalane}\:{constant} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{G}−\frac{\pi{ln}\left(\mathrm{2}\right)}{\mathrm{4}} \\ $$

Commented by Shrodinger last updated on 11/Jun/24

thanks sir..

$${thanks}\:{sir}.. \\ $$

Commented by Shrodinger last updated on 11/Jun/24

Sir It is ∫_0 ^(4/π) ln(cosx)dx

$${Sir}\:{It}\:{is}\:\int_{\mathrm{0}} ^{\frac{\mathrm{4}}{\pi}} {ln}\left({cosx}\right){dx} \\ $$

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