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ln-cosx-dx-




Question Number 149150 by mathdanisur last updated on 03/Aug/21
∫ ln (cosx) dx = ?
ln(cosx)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...
K=xln(cosx)+xtanxdx=xln(cosx)+xn=1B2n22n(22n1)x2n1(2n)!dx=xln(cosx)+n=1B2n22n(22n1)(2n)!x2ndxsoit:I=xln(cosx)+n=1B2n22n(22n1)(2n+1)!x2n+1+CLepuissant
Commented by mathdanisur last updated on 03/Aug/21
Thank You Ser
ThankYouSer

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