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Question Number 174192 by Best1 last updated on 26/Jul/22

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosxe}^{\left({cosxdx}\right)} {dx}\:\:\:????? \\$$

Commented by Best1 last updated on 27/Jul/22

$${help} \\$$

Commented by mr W last updated on 27/Jul/22

$${the}\:{question}\:{is}\:{non}−{sense}. \\$$$${cosxe}^{\left({cosxdx}\right)} \:{is}\:{not}\:{a}\:{valid}\:{function}! \\$$

Commented by Best1 last updated on 28/Jul/22

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosxe}^{\left(\int{cosxdx}\right)} {dx}? \\$$

Commented by Best1 last updated on 28/Jul/22

$$\\$$

Commented by mr W last updated on 28/Jul/22

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosxe}^{\left(\int{cosxdx}\right)} {dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {cosxe}^{\left(\mathrm{sin}\:{x}+{C}\right)} {dx} \\$$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{\left(\mathrm{sin}\:{x}+{C}\right)} {d}\left(\mathrm{sin}\:{x}+{C}\right) \\$$$$=\left[{e}^{\left(\mathrm{sin}\:{x}+{C}\right)} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \\$$$$={e}^{\mathrm{1}+{C}} −{e}^{{C}} \\$$$$=\left({e}−\mathrm{1}\right){e}^{{C}} \\$$