What is the integral of e^{-x}sinh(x) ?

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

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integral of e^{-x}sinh(x)

\int e^{- x} sinh (x) d x

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

Solution steps

\int e^{- x} sinh (x) d x

Apply u-substitution

: \int - sinh (- ln (u)) d u

= \int - sinh (- ln (u)) d u

Take the constant out:

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

= - \int sinh (- ln (u)) d u

Apply Integration By Parts

: u sinh (- ln (u)) - \int - cosh (- ln (u)) d u

= - (u sinh (- ln (u)) - \int - cosh (- ln (u)) d u)

\int - cosh (- ln (u)) d u = - \frac{1}{2} (\frac{u ^{2}}{2} + ln ∣ u ∣)

= - (u sinh (- ln (u)) - (- \frac{1}{2} (\frac{u ^{2}}{2} + ln ∣ u ∣)))

Substitute back

u = e^{- x}

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

Simplify

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

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

Add a constant to the solution

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

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