pi-4-pi-4-e-x-sin-x-dx-

Question Number 81664 by zainal tanjung last updated on 14/Feb/20

\underset{- π / 4}{\int^{π / 4}} e^{- x} \sin x d x =

Commented by zainal tanjung last updated on 14/Feb/20

Commented by abdomathmax last updated on 14/Feb/20

I = I m (\int_{- \frac{π}{4}}^{\frac{π}{4}} e^{- x} e^{i x} d x) = I m (\int_{- \frac{π}{4}}^{\frac{π}{4}} e^{(- 1 + i) x} d x)

a n d \int_{- \frac{π}{4}}^{\frac{π}{4}} e^{(- 1 + i) x} d x = {[\frac{1}{- 1 + i} e^{(- 1 + i) x}]}_{- \frac{π}{4}}^{\frac{π}{4}}

= \frac{- 1}{1 - i} {e^{(- 1 + i) \frac{π}{4}} - e^{- (- 1 + i) \frac{π}{4}}}

= \frac{- 1 - i}{2} {e^{- \frac{π}{4}} (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}}) - e^{\frac{π}{4}} (\frac{1}{\sqrt{2}} - \frac{i}{\sqrt{2}})}

= - \frac{1 + i}{2} {\frac{1}{\sqrt{2}} (e^{- \frac{π}{4}} - e^{\frac{π}{4}}) + \frac{i}{\sqrt{2}} (e^{\frac{π}{4}} + e^{- \frac{π}{4}})}

= - \frac{1}{\sqrt{2}} (1 + i) {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + (e^{\frac{π}{4}} + e^{- \frac{π}{4}}) i}

= - \frac{1}{\sqrt{2}} {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + i (e^{\frac{π}{4}} + e^{- \frac{π}{4}}) + i (e^{- \frac{π}{4}} - e^{\frac{π}{4}})

- (e^{\frac{π}{4}} - e^{- \frac{π}{4}})} = - \frac{1}{\sqrt{2}} {2 e^{- \frac{π}{4}} - 2 e^{\frac{π}{4}}

+ i (2 e^{- \frac{π}{4}})} \Rightarrow I = 2 e^{- \frac{π}{4}}

Commented by Tony Lin last updated on 14/Feb/20

\int e^{- x} s i n x d x

= - e^{- x} s i n x + \int e^{- x} c o s x d x

= - e^{- x} s i n x + (- e^{- x} c o s x - \int e^{- x} s i n x d x)

\Rightarrow \int e^{- x} s i n x d x

= \frac{- e^{- x} (s i n x + c o s x)}{2} + c

\Rightarrow \int_{- \frac{π}{4}}^{\frac{π}{4}} e^{- x} s i n x d x

= {[\frac{- e^{- x} (s i n x + c o s x)}{2}]}_{- \frac{π}{4}}^{\frac{π}{4}}

= - \frac{e^{- \frac{π}{4}}}{\sqrt{2}}

Commented by mathmax by abdo last updated on 14/Feb/20

s o r r y \int_{- \frac{π}{4}}^{\frac{π}{4}} e^{(- 1 + i) x} d x = - \frac{1}{2 \sqrt{2}} (1 + i) {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + i (e^{\frac{π}{4}} + e^{- \frac{π}{4}})}

= - \frac{1}{2 \sqrt{2}} {e^{- \frac{π}{4}} - e^{\frac{π}{4}} + i (e^{\frac{π}{4}} + e^{- \frac{π}{4}}) + i (e^{- \frac{π}{4}} - e^{\frac{π}{4}}) - (e^{\frac{π}{4}} + e^{- \frac{π}{4}})} \Rightarrow

I = - \frac{1}{2 \sqrt{2}} (2 e^{- \frac{π}{4}}) = - \frac{1}{\sqrt{2}} e^{- \frac{π}{4}}

★ I = - \frac{1}{\sqrt{2}} e^{- \frac{π}{4}} ★