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Question Number 168429 by mathocean1 last updated on 10/Apr/22

$$calculate:$$$$P={\int }_1^{e^{\frac{\pi }{2}}}\frac{cos\left(lnx\right)}{x}dx$$

Answered by qaz last updated on 10/Apr/22

$${\int }_1^{{\mathrm{e}}^{\pi /2}}\frac{\cos \left(\mathrm{lnx}\right)}{\mathrm{x}}\mathrm{dx}=\mathrm{\Re }{\int }_1^{{\mathrm{e}}^{\pi /2}}{\mathrm{x}}^{\mathrm{i}-1}\mathrm{dx}=\mathrm{\Re }\frac{{\mathrm{x}}^{\mathrm{i}}}{\mathrm{i}}{\mid }_1^{{\mathrm{e}}^{\pi /2}}=\mathrm{\Re }(\frac{{\mathrm{e}}^{\frac{\pi }{2}\mathrm{i}}}{\mathrm{i}}-\frac{1}{\mathrm{i}})=1$$

Answered by FelipeLz last updated on 10/Apr/22

$$\ln \left(x\right)=u\to \frac{1}{x}dx=du$$$$P={\int }_1^{e^{\frac{\pi }{2}}}\frac{\cos \left(\ln \left(x\right)\right)}{x}dx={\int }_0^{\frac{\pi }{2}}\cos \left(u\right)du=\sin \left(u\right){\mid }_0^{\frac{\pi }{2}}=1$$

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