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

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

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integral of (e^{ax}+e^{-ax})^2

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

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

Solution steps

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

Rewrite using trig identities

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

Simplify

(2 cosh (a x))^{2} : 4 cosh^{2} (a x)

= \int 4 cosh^{2} (a x) d x

Take the constant out:

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

= 4 \cdot \int cosh^{2} (a x) d x

Apply u-substitution

: \int \frac{cosh ^{2} ( u )}{a} d u

= 4 \cdot \int \frac{cosh ^{2} ( u )}{a} d u

Take the constant out:

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

= 4 \cdot \frac{1}{a} \cdot \int cosh^{2} (u) d u

Rewrite using trig identities

= 4 \cdot \frac{1}{a} \cdot \int \frac{1 + cosh ( 2 u )}{2} d u

Take the constant out:

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

= 4 \cdot \frac{1}{a} \cdot \frac{1}{2} \cdot \int 1 + cosh (2 u) d u

Apply the Sum Rule:

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

= 4 \cdot \frac{1}{a} \cdot \frac{1}{2} (\int 1 d u + \int cosh (2 u) d u)

\int 1 d u = u

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

= 4 \cdot \frac{1}{a} \cdot \frac{1}{2} (u + \frac{1}{2} sinh (2 u))

Substitute back

u = a x

= 4 \cdot \frac{1}{a} \cdot \frac{1}{2} (a x + \frac{1}{2} sinh (2 a x))

Simplify

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

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

Add a constant to the solution

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

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