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

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

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

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

Commented by bemath last updated on 18/Aug/20

$$jooss....$$

Answered by bemath last updated on 18/Aug/20

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

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

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

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