Integration Questions

Question Number 126298 by bramlexs22 last updated on 19/Dec/20

$$\:\:\int_{\:\mathrm{0}} ^{\:\sqrt{\mathrm{ln}\:\left(\pi/\mathrm{2}\right)}} {xe}^{{x}^{\mathrm{2}} } \mathrm{sin}\:\left({e}^{{x}^{\mathrm{2}} } \right)\:{dx}\:? \\$$

Answered by liberty last updated on 19/Dec/20

$${let}\:{v}={e}^{{x}^{\mathrm{2}} } \wedge\:{dv}=\mathrm{2}{xe}^{{x}^{\mathrm{2}} } {dx}\rightarrow\begin{cases}{{x}=\mathrm{0}\rightarrow{v}=\mathrm{1}}\\{{x}=\sqrt{\mathrm{ln}\:\left(\frac{\pi}{\mathrm{2}}\right)}\rightarrow{v}=\frac{\pi}{\mathrm{2}}}\end{cases} \\$$$${V}=\int_{\mathrm{1}} ^{\:\pi/\mathrm{2}} \frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:{v}\:{dv}\:=−\frac{\mathrm{1}}{\mathrm{2}}\left[\:\mathrm{cos}\:{v}\:\right]_{\mathrm{1}} ^{\frac{\pi}{\mathrm{2}}} \\$$$${V}=−\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{0}−\mathrm{cos}\:\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{cos}\:\mathrm{1} \\$$

Answered by Olaf last updated on 19/Dec/20

$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{ln}\frac{\pi}{\mathrm{2}}}} {xe}^{{x}^{\mathrm{2}} } \mathrm{sin}\left({e}^{{x}^{\mathrm{2}} } \right){dx} \\$$$$\mathrm{I}\:=\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{ln}\frac{\pi}{\mathrm{2}}}} −\frac{\mathrm{1}}{\mathrm{2}}.\frac{{d}}{{dx}}\mathrm{cos}\left({e}^{{x}^{\mathrm{2}} } \right){dx} \\$$$$\mathrm{I}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{cos}\left({e}^{{x}^{\mathrm{2}} } \right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{ln}\frac{\pi}{\mathrm{2}}}} =\:\frac{\mathrm{cos}\left(\mathrm{1}\right)}{\mathrm{2}} \\$$

Answered by Dwaipayan Shikari last updated on 19/Dec/20

$${e}^{{x}^{\mathrm{2}} } ={t}\:\:\:\: \\$$$$\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{1}} ^{\frac{\pi}{\mathrm{2}}} {sintdt}\:=\frac{\mathrm{1}}{\mathrm{2}}{cos}\left(\mathrm{1}\right) \\$$