using by parts integrate cos(log x)dx

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$$\begin{gathered} \mathrm{I}=\int \cos \left(\mathrm{logx}\right).\mathrm{dx} ;\left(\mathrm{i}\right) \\ =\int 1.\cos \left(\mathrm{logx}\right).\mathrm{dx} \\ \mathrm{Using} \mathrm{integration} \mathrm{by} \mathrm{part} \mathrm{taking} \cos \left(\mathrm{logx}\right) \mathrm{as} \mathrm{first} \mathrm{function} \mathrm{and} 1 \mathrm{as} \mathrm{second} \\ \mathrm{function}. \\ \Rightarrow \mathrm{I}=\cos \left(\mathrm{logx}\right)\int 1.\mathrm{dx}-\int \left[\frac{\mathrm{d}}{\mathrm{dx}}\left\{\cos \left(\mathrm{logx}\right)\right\}\int 1.\mathrm{dx}\right]\mathrm{dx} \\ =\cos \left(\mathrm{logx}\right).\mathrm{x}-\int \left\{-\sin \left(\mathrm{logx}\right).\frac{1}{\mathrm{x}}\right\}\mathrm{x}.\mathrm{dx} \left[\frac{\mathrm{d}}{\mathrm{dx}}\left(\cos \left(\mathrm{logx}\right)\right)=-\sin \left(\mathrm{logx}\right).\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{logx}\right)=-\sin \left(\mathrm{logx}\right).\frac{1}{\mathrm{x}}\right] \\ =\cos \left(\mathrm{logx}\right).\mathrm{x}+\int \sin \left(\mathrm{logx}\right).\mathrm{dx} \\ \Rightarrow \mathrm{I}=\mathrm{x}.\cos \left(\mathrm{logx}\right)+\int 1.\sin \left(\mathrm{logx}\right).\mathrm{dx} \\ \mathrm{Again} \mathrm{using} \mathrm{integration} \mathrm{by} \mathrm{part} \mathrm{taking} \sin \left(\mathrm{logx}\right) \mathrm{as} \mathrm{first} \mathrm{function} \mathrm{and} 1 \mathrm{as} \mathrm{second} \\ \mathrm{function} \\ \Rightarrow \mathrm{I}=\mathrm{x}.\cos \left(\mathrm{logx}\right)+\sin \left(\mathrm{logx}\right)\int 1.\mathrm{dx}-\int \left[\frac{\mathrm{d}}{\mathrm{dx}}\left\{\sin \left(\mathrm{logx}\right)\right\}\int 1.\mathrm{dx}\right]\mathrm{dx} \\ \Rightarrow \mathrm{I}=\mathrm{x}.\cos \left(\mathrm{logx}\right)+\sin \left(\mathrm{logx}\right).\mathrm{x}-\int \left[\cos \left(\mathrm{logx}\right).\frac{1}{\mathrm{x}}.\mathrm{x}\right]\mathrm{dx} \\ \Rightarrow \mathrm{I}=\mathrm{x}.\cos \left(\mathrm{logx}\right)+\mathrm{x}.\sin \left(\mathrm{logx}\right)-\int \cos \left(\mathrm{logx}\right)\mathrm{dx} \\ \mathrm{Using} \left(\mathrm{i}\right) \mathrm{we} \mathrm{have} \mathrm{I}=\int \cos \left(\mathrm{logx}\right).\mathrm{dx} \\ \Rightarrow \mathrm{I}=\mathrm{x}.\cos \left(\mathrm{logx}\right)+\sin \left(\mathrm{logx}\right).\mathrm{x}-\mathrm{I} \\ \Rightarrow \mathrm{I}+\mathrm{I}=\mathrm{x}\left[\cos \left(\mathrm{logx}\right)+\sin \left(\mathrm{logx}\right)\right] \\ \Rightarrow 2\mathrm{I}=\mathrm{x}\left[\cos \left(\mathrm{logx}\right)+\sin \left(\mathrm{logx}\right)\right] \\ \Rightarrow \mathrm{I}=\frac{\mathrm{x}}{2}\left[\cos \left(\mathrm{logx}\right)+\sin \left(\mathrm{logx}\right)\right]+\mathrm{C}, \mathrm{where} \mathrm{C} \mathrm{is} \mathrm{integration} \mathrm{constant}. \\ \Rightarrow \mathbf{\int }\mathbf{cos}\left(\mathbf{logx}\right)\mathbf{.}\mathbf{dx}\mathbf{=}\frac{\mathbf{x}}{\mathbf{2}}\left[\mathbf{cos}\left(\mathbf{logx}\right)\mathbf{+}\mathbf{sin}\left(\mathbf{logx}\right)\right]\mathbf{+}\mathbf{C} \end{gathered}$$

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