# What is intsinh(x)cosh(x)dx?

Mar 8, 2018

$\frac{1}{4} \cos 2 h \left(x\right) + C$

#### Explanation:

$\int \sinh \left(x\right) \cdot \cosh \left(x\right) \cdot \mathrm{dx}$

=$\frac{1}{2} \int \sin 2 h \left(x\right) \cdot \mathrm{dx}$

=$\frac{1}{4} \cos 2 h \left(x\right) + C$

Mar 8, 2018

$\int \setminus \sinh x \cosh x \setminus \mathrm{dx} = \frac{1}{4} \setminus \cosh 2 x + C$

#### Explanation:

We seek:

$I = \int \setminus \sinh x \cosh x \setminus \mathrm{dx}$

Method 1

We can use the identity:

$\sinh 2 x \equiv 2 \sinh x \cosh x$

Which gives us:

$I = \int \setminus \frac{1}{2} \sinh 2 x \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{2} \setminus \frac{\cosh 2 x}{2} + C$

$\setminus \setminus = \frac{1}{4} \setminus \cosh 2 x + C$

Method 2

Using the definitions of $\sinh x$ and $\cosh x$

$I = \int \setminus \frac{{e}^{x} - {e}^{- x}}{2} \frac{{e}^{x} + {e}^{- x}}{2} \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{4} \setminus \int \setminus \left({e}^{x} - {e}^{- x}\right) \left({e}^{x} + {e}^{- x}\right) \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{4} \setminus \int \setminus {e}^{2 x} - {e}^{- 2 x} \setminus \mathrm{dx}$

$\setminus \setminus = \frac{1}{4} \setminus \left\{{e}^{2 x} / 2 + {e}^{- 2 x} / 2\right\} + C$

$\setminus \setminus = \frac{1}{4} \setminus \frac{{e}^{2 x} + {e}^{- 2 x}}{2} + C$

$\setminus \setminus = \frac{1}{4} \setminus \cosh 2 x + C$