What is $\int \frac{e^{i x} - e^{- i x}}{2} d x$ $int (e^(ix)-e^(-ix))/2dx$ ?

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What is $\int \frac{e^{i x} - e^{- i x}}{2} d x$ $int (e^(ix)-e^(-ix))/2dx$ ?

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Answer 1 · 2018-05-02T11:03:43

$\int \frac{e^{i x} - e^{- i x}}{2} d x = - i cos x + C$ $int (e^(ix) - e^(-ix))/2dx = -icosx + C$

Explanation:

This requires using somewhat obscure trig identities, results which ultimately comes from Euler's formula.

The identities are $e^{i x} = cos x + i sin x$ $e^(ix) = cosx + isinx$ and $e^{- i x} = cos x - i sin x$ $e^(-ix) = cosx - isinx$ .

This turns $\int \frac{e^{i x} - e^{- i x}}{2} d x$ $int (e^(ix) - e^(-ix))/2 dx$ into $\int \frac{(cos x + i sin x) - (cos x - i sin x)}{2} = \int \frac{2 i sin x}{2} = \int (i sin x)$ $int ((cosx + isinx) - (cosx - isinx))/2 = int (2isinx)/2 = int(isinx)$

The $i$ $i$ is a constant, so this yields $i \int sin x d x = - i cos x + C$ $i int sinxdx = -i cosx + C$ , where $C$ $C$ is our constant of integration.

Answer 2 · 2018-05-02T11:04:03

See below

Explanation:

We can use the identity $sinh x = \frac{1}{2} \cdot (e^{x} - e^{- x})$ $sinhx=1/2·(e^x-e^(-x))$ where $sinh x$ $sinhx$ is the hyperbolic sin of x

in our case $sinh (i x)$ $sinh(ix)$ we know that $\int sinh x = cosh x$ $intsinhx=coshx$

Then $\int sinh i x d x = \frac{1}{i} cosh i x + C = - i cosh i x + C$ $intsinhixdx=1/icoshix+C=-icoshix+C$

Other way is

$\frac{1}{2} [\int e^{i x} d x - \int e^{- i x} d x] = \frac{1}{2} \cdot \frac{1}{i} e^{i x} + \frac{1}{2} \cdot \frac{1}{i} e^{- i x} = \frac{1}{i} cosh i x = - i cosh i x + C$ $1/2[inte^(ix)dx-inte^(-ix)dx]=1/2·1/ie^(ix)+1/2·1/ie^(-ix)=1/icoshix=-icoshix+C$

Note that $cosh x = \frac{1}{2} \cdot (e^{x} + e^{- x})$ $coshx=1/2·(e^x+e^(-x))$

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What is $\int \frac{e^{i x} - e^{- i x}}{2} d x$ $int (e^(ix)-e^(-ix))/2dx$ ?

2 Answers

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