How do I show this hyperbolic integral: $ \int_0^1 \sinh(2x)dx= \frac14 (e-\frac1e)^2$

Question

How do I show that: $$ \int_0^1 \sinh(2x)dx= \frac14 \left(e-\frac1e\right)^2$$

Answer 1

Note that $\sinh(2x) = 2\sinh x\cosh x$ and $(\sinh x)' =\cosh x$. Therefore $$ \int_0^1 \sinh 2x \,dx = \int_0^1 2\sinh x\cosh x \,dx = \int_0^1 (\sinh^2 x)'dx = \sinh^2 1 -\sinh^2 0 = (\frac{e-e^{-1}}{2})^2$$

Answer 2

Just note that $\sinh (2x) = \frac{e^{2x}-e^{-2x}}{2}$. The primitives are immediate.

$$ \int_0^1 \sinh (2x) dx = \frac 12 \int_0^1 (e^{2x}-e^{-2x}) dx = \frac 12 \left[\frac 12 e^{2x}+\frac 12 e^{-2x}\right]_0^1= \frac 14 (e^2+e^{-2}-2) = \frac 14 (e-e^{-1})^2 $$