What is the integral of sin(x)cosh(x) ?

The integral of sin(x)cosh(x) is 1/4 (-e^{-x}cos(x)-e^{-x}sin(x)-e^xcos(x)+e^xsin(x))+C

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integral of sin(x)cosh(x)

\int sin (x) cosh (x) d x

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

Solution steps

\int sin (x) cosh (x) d x

Rewrite using trig identities

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

Expand

sin (x) (e^{- x} + e^{x}) : e^{- x} sin (x) + e^{x} sin (x)

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

\int e^{- x} sin (x) d x = - \frac{e ^{- x} cos ( x )}{2} - \frac{e ^{- x} sin ( x )}{2}

\int e^{x} sin (x) d x = - \frac{e ^{x} cos ( x )}{2} + \frac{e ^{x} sin ( x )}{2}

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

Simplify

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

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

Add a constant to the solution

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

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