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

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

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

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

\frac{1}{2} (x + \frac{1}{2} sinh (2 x)) + C

Solution steps

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

Rewrite using trig identities

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

Simplify

= \int cosh^{2} (x) d x

Rewrite using trig identities

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

\int 1 d x = x

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

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

Add a constant to the solution

= \frac{1}{2} (x + \frac{1}{2} sinh (2 x)) + C

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