Question Number 35677 by abdo imad last updated on 21/May/18
$${find}\:{F}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} \:{e}^{−\mathrm{2}{t}} {cos}\left({t}+\frac{\pi}{\mathrm{4}}\right){dx}. \\ $$
Commented by prof Abdo imad last updated on 23/May/18
![we have F(x)= Re( ∫_0 ^x e^(−2t) e^(i(t+(π/4))) dt) =Re( ∫_0 ^x e^(−2t +i(t +(π/4))) dt) but ∫_0 ^x e^(−2t +i(t+(π/4))) dt = ∫_0 ^x e^((−2+i)t +i(π/4)) dt =e^(i(π/4)) ∫_0 ^x e^((−2+i)t) dt = e^(i(π/4)) [(1/(−2+i)) e^((−2+i)t) ]_0 ^x = (e^(i(π/4)) /(−2+i)){ e^((−2+i)x) −1} = (((−2−i) e^(i(π/4)) )/5) e^(−2x) { cosx +isinx −1} =−(1/5){ (2+i)(((√2)/2) +i((√2)/2))}e^(−2x) { cosx +isinx −1} =−((2(√2))/5){ (2+i)(1+i)}e^(−2x) {cosx +isinx−1} =−((2(√2))/5)e^(−2x) ( 2 +2i +i−1){ cosx +isinx −1} =−((2(√2))/5)e^(−2x) ( 1+3i){ cosx +isinx −1} = −((2(√2))/5) e^(−2x) { cosx +isinx −1 +3i cosx −3sinx−3i} F(x)= −((2(√2))/5) e^(−2x) { cosx −3sinx −1} .](https://www.tinkutara.com/question/Q35800.png)
$${we}\:{have}\:{F}\left({x}\right)=\:{Re}\left(\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−\mathrm{2}{t}} \:{e}^{{i}\left({t}+\frac{\pi}{\mathrm{4}}\right)} {dt}\right) \\ $$$$={Re}\left(\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−\mathrm{2}{t}\:+{i}\left({t}\:+\frac{\pi}{\mathrm{4}}\right)} \:{dt}\right)\:\:{but} \\ $$$$\int_{\mathrm{0}} ^{{x}} \:\:{e}^{−\mathrm{2}{t}\:+{i}\left({t}+\frac{\pi}{\mathrm{4}}\right)} {dt}\:=\:\int_{\mathrm{0}} ^{{x}} \:\:{e}^{\left(−\mathrm{2}+{i}\right){t}\:\:+{i}\frac{\pi}{\mathrm{4}}} \:{dt} \\ $$$$={e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\int_{\mathrm{0}} ^{{x}} \:\:{e}^{\left(−\mathrm{2}+{i}\right){t}} {dt}\:=\:{e}^{{i}\frac{\pi}{\mathrm{4}}} \:\:\left[\frac{\mathrm{1}}{−\mathrm{2}+{i}}\:{e}^{\left(−\mathrm{2}+{i}\right){t}} \right]_{\mathrm{0}} ^{{x}} \\ $$$$=\:\frac{{e}^{{i}\frac{\pi}{\mathrm{4}}} }{−\mathrm{2}+{i}}\left\{\:\:{e}^{\left(−\mathrm{2}+{i}\right){x}} \:−\mathrm{1}\right\} \\ $$$$=\:\frac{\left(−\mathrm{2}−{i}\right)\:{e}^{{i}\frac{\pi}{\mathrm{4}}} }{\mathrm{5}}\:{e}^{−\mathrm{2}{x}} \left\{\:{cosx}\:+{isinx}\:−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}}\left\{\:\left(\mathrm{2}+{i}\right)\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)\right\}{e}^{−\mathrm{2}{x}} \left\{\:{cosx}\:+{isinx}\:−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{5}}\left\{\:\left(\mathrm{2}+{i}\right)\left(\mathrm{1}+{i}\right)\right\}{e}^{−\mathrm{2}{x}} \left\{{cosx}\:+{isinx}−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{5}}{e}^{−\mathrm{2}{x}} \left(\:\mathrm{2}\:+\mathrm{2}{i}\:+{i}−\mathrm{1}\right)\left\{\:{cosx}\:+{isinx}\:−\mathrm{1}\right\} \\ $$$$=−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{5}}{e}^{−\mathrm{2}{x}} \:\left(\:\mathrm{1}+\mathrm{3}{i}\right)\left\{\:{cosx}\:+{isinx}\:−\mathrm{1}\right\} \\ $$$$=\:−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{5}}\:{e}^{−\mathrm{2}{x}} \left\{\:{cosx}\:+{isinx}\:−\mathrm{1}\:+\mathrm{3}{i}\:{cosx}\:−\mathrm{3}{sinx}−\mathrm{3}{i}\right\} \\ $$$${F}\left({x}\right)=\:−\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{5}}\:{e}^{−\mathrm{2}{x}} \left\{\:{cosx}\:−\mathrm{3}{sinx}\:−\mathrm{1}\right\}\:. \\ $$
Commented by prof Abdo imad last updated on 23/May/18
$${F}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{{x}} \:{e}^{−\mathrm{2}{t}} \:{cos}\left({t}+\frac{\pi}{\mathrm{4}}\right){dt}\:. \\ $$