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ln-cosx-dx-




Question Number 149150 by mathdanisur last updated on 03/Aug/21
∫ ln (cosx) dx = ?
$$\int\:{ln}\:\left({cosx}\right)\:{dx}\:=\:? \\ $$
Answered by puissant last updated on 03/Aug/21
K=xln(cosx)+∫xtanxdx  =xln(cosx)+∫xΣ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1)x^(2n−1) )/((2n)!))dx  =xln(cosx)+Σ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1))/((2n)!))∫x^(2n) dx  soit:  I=xln(cosx)+Σ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1))/((2n+1)!))x^(2n+1) +C               ......Le puissant...
$$\mathrm{K}=\mathrm{xln}\left(\mathrm{cosx}\right)+\int\mathrm{xtanxdx} \\ $$$$=\mathrm{xln}\left(\mathrm{cosx}\right)+\int\mathrm{x}\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\mathrm{B}_{\mathrm{2n}} \mid\frac{\mathrm{2}^{\mathrm{2n}} \left(\mathrm{2}^{\mathrm{2n}} −\mathrm{1}\right)\mathrm{x}^{\mathrm{2n}−\mathrm{1}} }{\left(\mathrm{2n}\right)!}\mathrm{dx} \\ $$$$=\mathrm{xln}\left(\mathrm{cosx}\right)+\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\mathrm{B}_{\mathrm{2n}} \mid\frac{\mathrm{2}^{\mathrm{2n}} \left(\mathrm{2}^{\mathrm{2n}} −\mathrm{1}\right)}{\left(\mathrm{2n}\right)!}\int\mathrm{x}^{\mathrm{2n}} \mathrm{dx} \\ $$$$\mathrm{soit}: \\ $$$$\mathrm{I}=\mathrm{xln}\left(\mathrm{cosx}\right)+\underset{\mathrm{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid\mathrm{B}_{\mathrm{2n}} \mid\frac{\mathrm{2}^{\mathrm{2n}} \left(\mathrm{2}^{\mathrm{2n}} −\mathrm{1}\right)}{\left(\mathrm{2n}+\mathrm{1}\right)!}\mathrm{x}^{\mathrm{2n}+\mathrm{1}} +\mathrm{C} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:……\mathrm{Le}\:\mathrm{puissant}… \\ $$
Commented by mathdanisur last updated on 03/Aug/21
Thank You Ser
$${Thank}\:{You}\:{Ser} \\ $$

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