What is the integral of (e^{-x}cosh(x))/2 ?

The integral of (e^{-x}cosh(x))/2 is 1/4 (-1/2 e^{-2x}+x)+C

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integral of (e^{-x}cosh(x))/2

\int \frac{e ^{- x} cosh ( x )}{2} d x

\frac{1}{4} (- \frac{1}{2} e^{- 2 x} + x) + C

Solution steps

\int \frac{e ^{- x} cosh ( x )}{2} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int e^{- x} cosh (x) d x

Rewrite using trig identities

= \frac{1}{2} \cdot \int \frac{1}{2} e^{- x} (e^{- x} + e^{x}) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int e^{- x} (e^{- x} + e^{x}) d x

Expand

e^{- x} (e^{- x} + e^{x}) : e^{- 2 x} + 1

= \frac{1}{2} \cdot \frac{1}{2} \cdot \int e^{- 2 x} + 1 d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{2} \cdot \frac{1}{2} (\int e^{- 2 x} d x + \int 1 d x)

\int e^{- 2 x} d x = - \frac{1}{2} e^{- 2 x}

\int 1 d x = x

= \frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{2} e^{- 2 x} + x)

Simplify

\frac{1}{2} \cdot \frac{1}{2} (- \frac{1}{2} e^{- 2 x} + x) : \frac{1}{4} (- \frac{1}{2} e^{- 2 x} + x)

= \frac{1}{4} (- \frac{1}{2} e^{- 2 x} + x)

Add a constant to the solution

= \frac{1}{4} (- \frac{1}{2} e^{- 2 x} + x) + C

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What is the integral of (e^{-x}cosh(x))/2 ?
The integral of (e^{-x}cosh(x))/2 is 1/4 (-1/2 e^{-2x}+x)+C