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sin-lnx-dx-




Question Number 120461 by Khalmohmmad last updated on 31/Oct/20
∫sin(lnx)dx
$$\int\mathrm{sin}\left(\mathrm{ln}{x}\right){dx} \\ $$
Answered by Dwaipayan Shikari last updated on 31/Oct/20
∫sin(logx)dx  =(∫e^t sin(t)dt )→I        logx=t  =−e^t cos(t)+∫e^t cost  =−e^t cos(t)+e^t sint−(∫e^t sintdt)→I  2I=e^t (sint−cost)  I=e^t (((sint)/2)−((cost)/2))  I=(x/2)(sin(logx)−cos(logx))+C  =(x/( (√2)))(sin(logx−(π/4)))+C
$$\int{sin}\left({logx}\right){dx} \\ $$$$=\left(\int{e}^{{t}} {sin}\left({t}\right){dt}\:\right)\rightarrow{I}\:\:\:\:\:\:\:\:{logx}={t} \\ $$$$=−{e}^{{t}} {cos}\left({t}\right)+\int{e}^{{t}} {cost} \\ $$$$=−{e}^{{t}} {cos}\left({t}\right)+{e}^{{t}} {sint}−\left(\int{e}^{{t}} {sintdt}\right)\rightarrow{I} \\ $$$$\mathrm{2}{I}={e}^{{t}} \left({sint}−{cost}\right) \\ $$$${I}={e}^{{t}} \left(\frac{{sint}}{\mathrm{2}}−\frac{{cost}}{\mathrm{2}}\right) \\ $$$${I}=\frac{{x}}{\mathrm{2}}\left({sin}\left({logx}\right)−{cos}\left({logx}\right)\right)+{C} \\ $$$$=\frac{{x}}{\:\sqrt{\mathrm{2}}}\left({sin}\left({logx}−\frac{\pi}{\mathrm{4}}\right)\right)+{C} \\ $$

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