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

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

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integral of 1/2 (e^x+e^{-x})

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

sinh (x) + C

Solution steps

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

Rewrite using trig identities

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

Use the common integral:

\int cosh (x) d x = sinh (x)

= \frac{1}{2} \cdot 2 sinh (x)

Simplify

\frac{1}{2} \cdot 2 sinh (x) : sinh (x)

= sinh (x)

Add a constant to the solution

= sinh (x) + C

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