Question Number 145776 by Engr_Jidda last updated on 08/Jul/21
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{x}} {cosxdx} \\ $$
Answered by ArielVyny last updated on 08/Jul/21
![I=∫_0 ^(π/2) e^x cosxdx du=e^x →u=e^x v=cosx→dv=−sinx I=[e^x cosx]_0 ^(π/2) +J (J=∫_0 ^(π/2) e^x sinxdx) J=[e^x sinx]_0 ^(π/2) −I { ((I−J=−1)),((I+J=e^(π/2) )) :} 2I=e^(π/2) −1→I=((e^(π/2) −1)/2)](https://www.tinkutara.com/question/Q145788.png)
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{x}} {cosxdx} \\ $$$${du}={e}^{{x}} \rightarrow{u}={e}^{{x}} \\ $$$${v}={cosx}\rightarrow{dv}=−{sinx} \\ $$$${I}=\left[{e}^{{x}} {cosx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +{J}\:\:\left({J}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {e}^{{x}} {sinxdx}\right) \\ $$$${J}=\left[{e}^{{x}} {sinx}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} −{I} \\ $$$$\begin{cases}{{I}−{J}=−\mathrm{1}}\\{{I}+{J}={e}^{\frac{\pi}{\mathrm{2}}} }\end{cases} \\ $$$$\mathrm{2}{I}={e}^{\frac{\pi}{\mathrm{2}}} −\mathrm{1}\rightarrow{I}=\frac{{e}^{\frac{\pi}{\mathrm{2}}} −\mathrm{1}}{\mathrm{2}} \\ $$$$ \\ $$
Answered by mathmax by abdo last updated on 08/Jul/21
![∫_0 ^(π/2) e^x cosx dx =Re(∫_0 ^(π/2) e^(x+ix) dx)=Re(∫_0 ^(π/2) e^((1+i)x) dx) and ∫_0 ^(π/2) e^((1+i)x) dx =[(1/(1+i))e^((1+i)x) ]_0 ^(π/2) =(1/(1+i))(e^((1+i)(π/2)) −1) =((1−i)/2)(i e^(π/2) −1) =((ie^(π/2) −1+e^(π/2) +i)/2) =((e^(π/2) −1)/2)+i(...) ⇒ ∫_0 ^(π/2) e^x cosxdx=((e^(π/2) −1)/2)](https://www.tinkutara.com/question/Q145789.png)
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{\mathrm{x}} \:\mathrm{cosx}\:\mathrm{dx}\:=\mathrm{Re}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{\mathrm{x}+\mathrm{ix}} \mathrm{dx}\right)=\mathrm{Re}\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \mathrm{dx}\right)\:\mathrm{and} \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \:\mathrm{dx}\:=\left[\frac{\mathrm{1}}{\mathrm{1}+\mathrm{i}}\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\mathrm{x}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:=\frac{\mathrm{1}}{\mathrm{1}+\mathrm{i}}\left(\mathrm{e}^{\left(\mathrm{1}+\mathrm{i}\right)\frac{\pi}{\mathrm{2}}} −\mathrm{1}\right) \\ $$$$=\frac{\mathrm{1}−\mathrm{i}}{\mathrm{2}}\left(\mathrm{i}\:\mathrm{e}^{\frac{\pi}{\mathrm{2}}} −\mathrm{1}\right)\:=\frac{\mathrm{ie}^{\frac{\pi}{\mathrm{2}}} −\mathrm{1}+\mathrm{e}^{\frac{\pi}{\mathrm{2}}} \:+\mathrm{i}}{\mathrm{2}}\:=\frac{\mathrm{e}^{\frac{\pi}{\mathrm{2}}} −\mathrm{1}}{\mathrm{2}}+\mathrm{i}\left(…\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\mathrm{e}^{\mathrm{x}} \mathrm{cosxdx}=\frac{\mathrm{e}^{\frac{\pi}{\mathrm{2}}} −\mathrm{1}}{\mathrm{2}} \\ $$