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

The integral of (e^t-e^{-t})^2 is -2t+sinh(2t)+C

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integral of (e^t-e^{-t})^2

\int (e^{t} - e^{- t})^{2} d t

- 2 t + sinh (2 t) + C

Solution steps

\int (e^{t} - e^{- t})^{2} d t

Rewrite using trig identities

= \int (2 sinh (t))^{2} d t

Simplify

(2 sinh (t))^{2} : 4 sinh^{2} (t)

= \int 4 sinh^{2} (t) d t

Take the constant out:

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

= 4 \cdot \int sinh^{2} (t) d t

Rewrite using trig identities

= 4 \cdot \int \frac{- 1 + cosh ( 2 t )}{2} d t

Take the constant out:

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

= 4 \cdot \frac{1}{2} \cdot \int - 1 + cosh (2 t) d t

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}{2} (- \int 1 d t + \int cosh (2 t) d t)

\int 1 d t = t

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

= 4 \cdot \frac{1}{2} (- t + \frac{1}{2} sinh (2 t))

Simplify

4 \cdot \frac{1}{2} (- t + \frac{1}{2} sinh (2 t)) : - 2 t + sinh (2 t)

= - 2 t + sinh (2 t)

Add a constant to the solution

= - 2 t + sinh (2 t) + C

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