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Question Number 145776 by Engr_Jidda last updated on 08/Jul/21

$${\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$$

Answered by ArielVyny last updated on 08/Jul/21

$$I={\int }_0^{\frac{\pi }{2}}{e}^{x}cosxdx$$$$du=e^x\to u=e^x$$$$v=cosx\to dv=-sinx$$$$I={\left[e^{x}cosx\right]}_0^{\frac{\pi }{2}}+J(J={\int }_0^{\frac{\pi }{2}}{e}^{x}sinxdx)$$$$J={\left[e^{x}sinx\right]}_0^{\frac{\pi }{2}}-I$$$$\{\begin{array}{l}I-J=-1\\ I+J=e^{\frac{\pi }{2}}\end{array}$$$$2I=e^{\frac{\pi }{2}}-1\to I=\frac{e^{\frac{\pi }{2}}-1}{2}$$$$$$

Answered by mathmax by abdo last updated on 08/Jul/21

$${\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{\mathrm{x}}\mathrm{cosx}\mathrm{dx}=\mathrm{Re}\left({\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{\mathrm{x}+\mathrm{ix}}\mathrm{dx}\right)=\mathrm{Re}\left({\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{(1+\mathrm{i})\mathrm{x}}\mathrm{dx}\right)\mathrm{and}$$$${\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{(1+\mathrm{i})\mathrm{x}}\mathrm{dx}={\left[\frac{1}{1+\mathrm{i}}{\mathrm{e}}^{(1+\mathrm{i})\mathrm{x}}\right]}_0^{\frac{\pi }{2}}=\frac{1}{1+\mathrm{i}}({\mathrm{e}}^{(1+\mathrm{i})\frac{\pi }{2}}-1)$$$$=\frac{1-\mathrm{i}}{2}(\mathrm{i}{\mathrm{e}}^{\frac{\pi }{2}}-1)=\frac{{\mathrm{ie}}^{\frac{\pi }{2}}-1+{\mathrm{e}}^{\frac{\pi }{2}}+\mathrm{i}}{2}=\frac{{\mathrm{e}}^{\frac{\pi }{2}}-1}{2}+\mathrm{i}(...)\Rightarrow$$$${\int }_0^{\frac{\pi }{2}}{\mathrm{e}}^{\mathrm{x}}\mathrm{cosxdx}=\frac{{\mathrm{e}}^{\frac{\pi }{2}}-1}{2}$$

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