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Question Number 173500 by Shrinava last updated on 12/Jul/22

$$\mathrm{Find}\mathrm{without}\mathrm{any}\mathrm{software}:$$$$\mathrm{\Omega }=\int (\mathrm{x}+\frac{5}{\mathrm{x}})(1-\frac{5}{{\mathrm{x}}^2})\sin \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\mathrm{dx}$$

Answered by thfchristopher last updated on 12/Jul/22

$$\mathrm{Let}u=x+\frac{5}{x}$$$$du=(1-\frac{5}{x^2})dx$$$$\therefore \mathrm{\Omega }=\int u\sin \left(\ln u\right)du$$$$=\frac{1}{2}\int \sin \left(\ln u\right)d\left(u^2\right)$$$$=\frac{1}{2}[u^2\sin \left(\ln u\right)-\int u^{2}d\left\{\sin \left(\ln u\right)\right\}]$$$$=\frac{1}{2}{u}^2\sin \left(\ln u\right)-\frac{1}{2}\int u\cos \left(\ln u\right)du$$$$=\frac{1}{2}{u}^2\sin \left(\ln u\right)-\frac{1}{4}\int \cos \left(\ln u\right)d\left(u^2\right)$$$$=\frac{1}{2}{u}^2\sin \left(\ln u\right)-\frac{1}{4}[u^2\cos \left(\ln u\right)-\int u^{2}d\left\{\cos \left(\ln u\right)\right\}]$$$$=\frac{1}{2}{u}^2\sin \left(\ln u\right)-\frac{1}{4}{u}^2\cos \left(\ln u\right)-\frac{1}{4}\int u\sin \left(\ln u\right)du$$$$\Rightarrow \frac{5}{4}\int u\sin \left(\ln u\right)du=\frac{1}{4}{u}^2[2\sin \left(\ln u\right)-\cos \left(\ln u\right)]+C$$$$\int u\sin \left(\ln u\right)du=\frac{1}{5}{u}^2[2\sin \left(\ln u\right)-\cos \left(\ln u\right)]+C$$$$\Rightarrow \int (x+\frac{5}{x})(1-\frac{5}{x^2})\sin \left[\ln (x+\frac{5}{x})\right]dx$$$$=\frac{1}{5}{(x+\frac{5}{x})}^2[2\sin \left\{\ln (x+\frac{5}{x})\right\}-\cos \left\{\ln (x+\frac{5}{x})\right\}]+C$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}.$$

Answered by a.lgnaoui last updated on 13/Jul/22

$$1-\frac{5}{{\mathrm{x}}^2}={(\mathrm{x}+\frac{5}{\mathrm{x}})}^{\prime }and\frac{{(\mathrm{x}+\frac{5}{\mathrm{x}})}^{\prime }}{x+\frac{5}{\mathrm{x}}}=\ln {(\mathrm{x}+\frac{5}{\mathrm{x}})}^{\prime }$$$$\rfloor \mathrm{\Omega }=\int {(x+\frac{5}{x})}^2\times \ln {(x+\frac{5}{x})}^{\prime }\sin \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\mathrm{dx}$$$$=\int {(x+\frac{5}{x})}^2\times [-\cos {\left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right]}^{\prime }dx$$$$=let\mathrm{U}={(x+\frac{5}{x})}^{2}and\mathrm{V}=-\cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)$$$$\mathrm{\Omega }=\int \mathrm{U}\times {\mathrm{V}}^{\prime }=\mathrm{UV}-\int {\mathrm{U}}^{\prime }\mathrm{V}$$$$[-{(x+\frac{5}{x})}^2\times \cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)]+\int {\left[{(x+\frac{5}{x})}^2\right]}^{\prime }\cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\mathrm{dx}$$$$\text{Prime causes double exponent: use braces to clarify}$$$$posons\int {\mathrm{U}}^{\prime }\cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)=2\int (x+\frac{5}{\mathrm{x}})(1-\frac{5}{x^2})\cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\mathrm{dx}$$$$$$$$2\int {(x+\frac{5}{x})}^2{\left[\sin \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\right]}^{{}^{\prime }}\mathrm{dx}$$$$=-2\mathrm{I}or\mathrm{I}=[-\cos \left(\ln (x+\frac{5}{x})\right){(\mathrm{x}+\frac{5}{\mathrm{x}})}^2]-2\mathrm{I}$$$$3\mathrm{I}=-{(x+\frac{5}{x})}^2\cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\Rightarrow$$$$\mathrm{I}=-\frac{\left[{(x+\frac{5}{x})}^2\cos \left(\ln (\mathrm{x}+\frac{5}{\mathrm{x}})\right)\right]}{3}$$

Commented by Tawa11 last updated on 13/Jul/22

$$\mathrm{Great}\mathrm{sir}$$

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