ln-cosx-dx-

Question Number 145412 by Jamshidbek last updated on 04/Jul/21

$$\:\:\:\:\int\mathrm{ln}\left(\mathrm{cosx}\right)\mathrm{dx}=? \\ $$

Answered by Olaf_Thorendsen last updated on 04/Jul/21

F(x) = ∫ln(cosx) dx F(x) = xln(cosx)+∫xtanx dx F(x) = xln(cosx)+ ∫xΣ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1)x^(2n−1) )/((2n)!)) dx ∣x∣<(π/2) F(x) = xln(cosx)+ ∫Σ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1)x^(2n) )/((2n)!)) dx F(x) = xln(cosx)+Σ_(n=1) ^∞ ∣B_(2n) ∣((2^(2n) (2^(2n) −1)x^(2n+1) )/((2n+1)!))+C (many formulas possible but no explicit formula)

$$\mathrm{F}\left({x}\right)\:=\:\int\mathrm{ln}\left(\mathrm{cos}{x}\right)\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{ln}\left(\mathrm{cos}{x}\right)+\int{x}\mathrm{tan}{x}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{ln}\left(\mathrm{cos}{x}\right)+ \\ $$$$\int{x}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{B}_{\mathrm{2}{n}} \mid\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){x}^{\mathrm{2}{n}−\mathrm{1}} }{\left(\mathrm{2}{n}\right)!}\:{dx}\:\:\:\mid{x}\mid<\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{ln}\left(\mathrm{cos}{x}\right)+ \\ $$$$\int\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{B}_{\mathrm{2}{n}} \mid\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){x}^{\mathrm{2}{n}} }{\left(\mathrm{2}{n}\right)!}\:{dx} \\ $$$$\mathrm{F}\left({x}\right)\:=\:{x}\mathrm{ln}\left(\mathrm{cos}{x}\right)+\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\mid{B}_{\mathrm{2}{n}} \mid\frac{\mathrm{2}^{\mathrm{2}{n}} \left(\mathrm{2}^{\mathrm{2}{n}} −\mathrm{1}\right){x}^{\mathrm{2}{n}+\mathrm{1}} }{\left(\mathrm{2}{n}+\mathrm{1}\right)!}+\mathrm{C} \\ $$$$\left({many}\:{formulas}\:{possible}\right. \\ $$$$\left.{but}\:{no}\:{explicit}\:{formula}\right) \\ $$

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