0-pi-2-e-x-cosxdx-

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

\int_{0}^{\frac{π}{2}} e^{x} c o s x d x

Answered by ArielVyny last updated on 08/Jul/21

I = \int_{0}^{\frac{π}{2}} e^{x} c o s x d x

d u = e^{x} \to u = e^{x}

v = c o s x \to d v = - s i n x

I = {[e^{x} c o s x]}_{0}^{\frac{π}{2}} + J (J = \int_{0}^{\frac{π}{2}} e^{x} s i n x d x)

J = {[e^{x} s i n x]}_{0}^{\frac{π}{2}} - I

{\begin{cases} I - J = - 1 \\ I + J = e^{\frac{π}{2}} \end{cases}

2 I = e^{\frac{π}{2}} - 1 \to I = \frac{e^{\frac{π}{2}} - 1}{2}

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

\int_{0}^{\frac{π}{2}} e^{x} cosx dx = Re (\int_{0}^{\frac{π}{2}} e^{x + ix} dx) = Re (\int_{0}^{\frac{π}{2}} e^{(1 + i) x} dx) and

\int_{0}^{\frac{π}{2}} e^{(1 + i) x} dx = {[\frac{1}{1 + i} e^{(1 + i) x}]}_{0}^{\frac{π}{2}} = \frac{1}{1 + i} (e^{(1 + i) \frac{π}{2}} - 1)

= \frac{1 - i}{2} (i e^{\frac{π}{2}} - 1) = \frac{{ie}^{\frac{π}{2}} - 1 + e^{\frac{π}{2}} + i}{2} = \frac{e^{\frac{π}{2}} - 1}{2} + i (\dots) \Rightarrow

\int_{0}^{\frac{π}{2}} e^{x} cosxdx = \frac{e^{\frac{π}{2}} - 1}{2}