find F(x)=∫_0 ^x e^(−2t) cos(t+(π/4))dx.


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Question Number 35677 by abdo imad last updated on 21/May/18

$f i n d F (x) = \int_{0}^{x} e^{- 2 t} c o s (t + \frac{π}{4}) d x .$
Commented by prof Abdo imad last updated on 23/May/18
$we have F(x)= Re( ∫_0 ^x e^(−2t) e^(i(t+(π/4))) dt) =Re( ∫_0 ^x e^(−2t +i(t +(π/4))) dt) but ∫_0 ^x e^(−2t +i(t+(π/4))) dt = ∫_0 ^x e^((−2+i)t +i(π/4)) dt =e^(i(π/4)) ∫_0 ^x e^((−2+i)t) dt = e^(i(π/4)) [(1/(−2+i)) e^((−2+i)t) ]_0 ^x = (e^(i(π/4)) /(−2+i)){ e^((−2+i)x) −1} = (((−2−i) e^(i(π/4)) )/5) e^(−2x) { cosx +isinx −1} =−(1/5){ (2+i)(((√2)/2) +i((√2)/2))}e^(−2x) { cosx +isinx −1} =−((2(√2))/5){ (2+i)(1+i)}e^(−2x) {cosx +isinx−1} =−((2(√2))/5)e^(−2x) ( 2 +2i +i−1){ cosx +isinx −1} =−((2(√2))/5)e^(−2x) ( 1+3i){ cosx +isinx −1} = −((2(√2))/5) e^(−2x) { cosx +isinx −1 +3i cosx −3sinx−3i} F(x)= −((2(√2))/5) e^(−2x) { cosx −3sinx −1} .$
$w e h a v e F (x) = R e (\int_{0}^{x} e^{- 2 t} e^{i (t + \frac{π}{4})} d t)$ $= R e (\int_{0}^{x} e^{- 2 t + i (t + \frac{π}{4})} d t) b u t$ $\int_{0}^{x} e^{- 2 t + i (t + \frac{π}{4})} d t = \int_{0}^{x} e^{(- 2 + i) t + i \frac{π}{4}} d t$ $= e^{i \frac{π}{4}} \int_{0}^{x} e^{(- 2 + i) t} d t = e^{i \frac{π}{4}} {[\frac{1}{- 2 + i} e^{(- 2 + i) t}]}_{0}^{x}$ $= \frac{e^{i \frac{π}{4}}}{- 2 + i} {e^{(- 2 + i) x} - 1}$ $= \frac{(- 2 - i) e^{i \frac{π}{4}}}{5} e^{- 2 x} {c o s x + i s i n x - 1}$ $= - \frac{1}{5} {(2 + i) (\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2})} e^{- 2 x} {c o s x + i s i n x - 1}$ $= - \frac{2 \sqrt{2}}{5} {(2 + i) (1 + i)} e^{- 2 x} {c o s x + i s i n x - 1}$ $= - \frac{2 \sqrt{2}}{5} e^{- 2 x} (2 + 2 i + i - 1) {c o s x + i s i n x - 1}$ $= - \frac{2 \sqrt{2}}{5} e^{- 2 x} (1 + 3 i) {c o s x + i s i n x - 1}$ $= - \frac{2 \sqrt{2}}{5} e^{- 2 x} {c o s x + i s i n x - 1 + 3 i c o s x - 3 s i n x - 3 i}$ $F (x) = - \frac{2 \sqrt{2}}{5} e^{- 2 x} {c o s x - 3 s i n x - 1} .$
Commented by prof Abdo imad last updated on 23/May/18

$F (x) = \int_{0}^{x} e^{- 2 t} c o s (t + \frac{π}{4}) d t .$
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