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Question Number 108597 by bemath last updated on 18/Aug/20

((∠ BeMath∠)/∦) (1) ∫ cos (ln x) dx (2) ∫ sin (ln x) dx

$$\:\:\:\:\:\frac{\angle\:\mathcal{B}{e}\mathcal{M}{ath}\angle}{\nparallel} \\ $$$$\left(\mathrm{1}\right)\:\int\:\mathrm{cos}\:\left(\mathrm{ln}\:{x}\right)\:{dx}\: \\ $$$$\left(\mathrm{2}\right)\:\int\:\mathrm{sin}\:\left(\mathrm{ln}\:{x}\right)\:{dx}\: \\ $$

Answered by 1549442205PVT last updated on 18/Aug/20

Set I=∫cos(lnx)dx,J=∫sin(lnx)dx By integrating by parts we have I=xcos(lnx)−∫x.(−sin(lnx)).(1/x)dx =xcos(lnx)+J⇒I−J=xcos(lnx)(1) J=∫sin(lnx)dx=xsin(lnx)−∫x.cos(lnx).(1/x)dx =xsin(lnx)−J⇒I+J=xsin(lnx)(2) Solve the system of equations (1)(2) we get { ((I=∫cos(lnx)dx=((x[(cos(lnx)+sin(lnx)])/2)+C)),((J=∫sin(lnx)dx=((x[sin(lnx)−cos(lnx)])/2)+C )) :}

$$\mathrm{Set}\:\mathrm{I}=\int\mathrm{cos}\left(\mathrm{lnx}\right)\mathrm{dx},\mathrm{J}=\int\mathrm{sin}\left(\mathrm{lnx}\right)\mathrm{dx} \\ $$$$\mathrm{By}\:\mathrm{integrating}\:\mathrm{by}\:\mathrm{parts}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{I}=\mathrm{xcos}\left(\mathrm{lnx}\right)−\int\mathrm{x}.\left(−\mathrm{sin}\left(\mathrm{lnx}\right)\right).\frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{xcos}\left(\mathrm{lnx}\right)+\mathrm{J}\Rightarrow\mathrm{I}−\mathrm{J}=\mathrm{xcos}\left(\mathrm{lnx}\right)\left(\mathrm{1}\right) \\ $$$$\mathrm{J}=\int\mathrm{sin}\left(\mathrm{lnx}\right)\mathrm{dx}=\mathrm{xsin}\left(\mathrm{lnx}\right)−\int\mathrm{x}.\mathrm{cos}\left(\mathrm{lnx}\right).\frac{\mathrm{1}}{\mathrm{x}}\mathrm{dx} \\ $$$$=\mathrm{xsin}\left(\mathrm{lnx}\right)−\mathrm{J}\Rightarrow\mathrm{I}+\mathrm{J}=\mathrm{xsin}\left(\mathrm{lnx}\right)\left(\mathrm{2}\right) \\ $$$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system}\:\mathrm{of}\:\mathrm{equations}\:\left(\mathrm{1}\right)\left(\mathrm{2}\right) \\ $$$$\mathrm{we}\:\mathrm{get} \\ $$$$\begin{cases}{\boldsymbol{\mathrm{I}}=\int\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{lnx}}\right)\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{x}}\left[\left(\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{lnx}}\right)+\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{lnx}}\right)\right]\right.}{\mathrm{2}}+\mathrm{C}}\\{\boldsymbol{\mathrm{J}}=\int\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{lnx}}\right)\boldsymbol{\mathrm{dx}}=\frac{\boldsymbol{\mathrm{x}}\left[\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{lnx}}\right)−\boldsymbol{\mathrm{cos}}\left(\boldsymbol{\mathrm{lnx}}\right)\right]}{\mathrm{2}}+\mathrm{C}\:}\end{cases} \\ $$

Commented by bemath last updated on 18/Aug/20

jooss....

$${jooss}.... \\ $$

Answered by bemath last updated on 18/Aug/20

(1) set ln x = q ⇒x = e^q ; dx =e^q dq I=∫ e^q cos q dq = ∫e^q d(sin q) = e^q sin q +∫ e^q d(cos q) =e^q sin q + e^q cos q −I 2I = e^q (sin q + cos q ) I=((x(sin (ln x)+cos (ln x)))/2)+C

$$\left(\mathrm{1}\right)\:{set}\:\mathrm{ln}\:{x}\:=\:{q}\:\Rightarrow{x}\:=\:{e}^{{q}} \:;\:{dx}\:={e}^{{q}} \:{dq} \\ $$$${I}=\int\:{e}^{{q}} \:\mathrm{cos}\:{q}\:{dq}\:=\:\int{e}^{{q}} \:{d}\left(\mathrm{sin}\:{q}\right) \\ $$$$=\:{e}^{{q}} \:\mathrm{sin}\:{q}\:+\int\:{e}^{{q}} \:{d}\left(\mathrm{cos}\:{q}\right)\: \\ $$$$={e}^{{q}} \:\mathrm{sin}\:{q}\:+\:{e}^{{q}} \:\mathrm{cos}\:{q}\:−{I} \\ $$$$\mathrm{2}{I}\:=\:{e}^{{q}} \left(\mathrm{sin}\:{q}\:+\:\mathrm{cos}\:{q}\:\right) \\ $$$${I}=\frac{{x}\left(\mathrm{sin}\:\left(\mathrm{ln}\:{x}\right)+\mathrm{cos}\:\left(\mathrm{ln}\:{x}\right)\right)}{\mathrm{2}}+{C} \\ $$

Answered by mathmax by abdo last updated on 18/Aug/20

1) I =∫ cos(lnx)dx changement lnx =t give x=e^t ⇒ I =∫ cost e^t dt =∫ e^t cost =_(byparts) e^t sint −∫ e^t sint dt =e^t sint −{ −e^t cost +∫ e^t cost dt} =e^t sint +e^t cost −I+c ⇒2I =(cost +sint)e^t +c ⇒I =(1/2)(cost +sint)e^t +C =(x/2)(cos(lnx)+sin(lnx)) +C

$$\left.\mathrm{1}\right)\:\mathrm{I}\:=\int\:\mathrm{cos}\left(\mathrm{lnx}\right)\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{lnx}\:=\mathrm{t}\:\mathrm{give}\:\mathrm{x}=\mathrm{e}^{\mathrm{t}} \:\Rightarrow \\ $$$$\mathrm{I}\:=\int\:\mathrm{cost}\:\mathrm{e}^{\mathrm{t}} \:\mathrm{dt}\:=\int\:\mathrm{e}^{\mathrm{t}} \:\mathrm{cost}\:=_{\mathrm{byparts}} \:\:\mathrm{e}^{\mathrm{t}} \mathrm{sint}\:−\int\:\mathrm{e}^{\mathrm{t}} \:\mathrm{sint}\:\mathrm{dt} \\ $$$$=\mathrm{e}^{\mathrm{t}} \:\mathrm{sint}\:−\left\{\:\:−\mathrm{e}^{\mathrm{t}} \mathrm{cost}\:+\int\:\mathrm{e}^{\mathrm{t}} \mathrm{cost}\:\mathrm{dt}\right\} \\ $$$$=\mathrm{e}^{\mathrm{t}} \mathrm{sint}\:\:+\mathrm{e}^{\mathrm{t}} \:\mathrm{cost}\:−\mathrm{I}+\mathrm{c}\:\Rightarrow\mathrm{2I}\:=\left(\mathrm{cost}\:+\mathrm{sint}\right)\mathrm{e}^{\mathrm{t}} +\mathrm{c}\:\Rightarrow\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{cost}\:+\mathrm{sint}\right)\mathrm{e}^{\mathrm{t}} \:+\mathrm{C} \\ $$$$=\frac{\mathrm{x}}{\mathrm{2}}\left(\mathrm{cos}\left(\mathrm{lnx}\right)+\mathrm{sin}\left(\mathrm{lnx}\right)\right)\:+\mathrm{C} \\ $$

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