Integration Questions

Question Number 149273 by john_santu last updated on 04/Aug/21

$$\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\frac{\mathrm{e}^{\mathrm{tan}\:\mathrm{x}} \:\mathrm{sin}\:^{\mathrm{2}} \mathrm{x}}{\mathrm{cos}\:^{\mathrm{4}} \mathrm{x}}\:\mathrm{dx}\:=? \\$$

Answered by Ar Brandon last updated on 04/Aug/21

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{e}^{\mathrm{tan}{x}} \mathrm{sin}^{\mathrm{2}} {x}}{\mathrm{cos}^{\mathrm{4}} {x}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {e}^{\mathrm{tan}{x}} \mathrm{sec}^{\mathrm{2}} {x}\centerdot\mathrm{tan}^{\mathrm{2}} {xdx} \\$$$$\:\:\:=\left[{e}^{\mathrm{tan}{x}} \centerdot\mathrm{tan}^{\mathrm{2}} {x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} −\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {e}^{\mathrm{tan}{x}} \mathrm{sec}^{\mathrm{2}} {x}\centerdot\mathrm{tan}{xdx} \\$$$$\:\:\:={e}−\mathrm{2}\left[{e}^{\mathrm{tan}{x}} \mathrm{tan}{x}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} +\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} {e}^{\mathrm{tan}{x}} \mathrm{sec}^{\mathrm{2}} {xdx} \\$$$$\:\:\:={e}−\mathrm{2}{e}+\mathrm{2}\left[{e}^{\mathrm{tan}{x}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} =−{e}+\mathrm{2}\left({e}−\mathrm{1}\right)={e}−\mathrm{2} \\$$

Commented by Ar Brandon last updated on 04/Aug/21

$$\mathrm{Integration}\:\mathrm{by}-\mathrm{part}\:\mathrm{with}\:{v}'\left({x}\right)={e}^{\mathrm{tan}{x}} \mathrm{sec}^{\mathrm{2}} {x} \\$$

Commented by puissant last updated on 30/Aug/21

$${hummm} \\$$

Answered by john_santu last updated on 04/Aug/21

Answered by ArielVyny last updated on 04/Aug/21

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{e}^{{tanx}} }{{cos}^{\mathrm{2}} {x}}×{tan}^{\mathrm{2}} {xdx} \\$$$${t}={tanx}\rightarrow{dt}=\frac{\mathrm{1}}{{cos}^{\mathrm{2}} {x}}{dx} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{t}} ×{t}^{\mathrm{2}} {dt}\rightarrow\begin{cases}{{du}={e}^{{t}} \rightarrow{u}={e}^{{t}} }\\{{v}={t}^{\mathrm{2}} \rightarrow{dv}=\mathrm{2}{t}}\end{cases} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} {e}^{{t}} {dt}=\left[{t}^{\mathrm{2}} {e}^{{t}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{t}} {t}={e}^{\mathrm{1}} −\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{t}} {tdt} \\$$$$\begin{cases}{{du}={e}^{{t}} \rightarrow{u}={e}^{{t}} }\\{{v}={t}\rightarrow{dv}=\mathrm{1}}\end{cases} \\$$$$\int_{\mathrm{0}} ^{\mathrm{1}} {te}^{{t}} {dt}=\left[{te}^{{t}} \right]_{\mathrm{0}} ^{\mathrm{1}} −\int_{\mathrm{0}} ^{\mathrm{1}} {e}^{{t}} {dt}=\left[{e}^{{t}} \left({t}−\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{1}} =\mathrm{1} \\$$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \frac{{e}^{{tanx}} {sin}^{\mathrm{2}} {x}}{{cos}^{\mathrm{4}} {x}}{dx}={e}^{\mathrm{1}} −\mathrm{2} \\$$