Integral of $$$\sinh^{2}{\left(x \right)}$$$
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Find $$$\int \sinh^{2}{\left(x \right)}\, dx$$$.
Solution
Apply the power reducing formula $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$ with $$$\alpha=x$$$:
$${\color{red}{\int{\sinh^{2}{\left(x \right)} d x}}} = {\color{red}{\int{\left(\frac{\cosh{\left(2 x \right)}}{2} - \frac{1}{2}\right)d x}}}$$
Apply the constant multiple rule $$$\int c f{\left(x \right)}\, dx = c \int f{\left(x \right)}\, dx$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(x \right)} = \cosh{\left(2 x \right)} - 1$$$:
$${\color{red}{\int{\left(\frac{\cosh{\left(2 x \right)}}{2} - \frac{1}{2}\right)d x}}} = {\color{red}{\left(\frac{\int{\left(\cosh{\left(2 x \right)} - 1\right)d x}}{2}\right)}}$$
Integrate term by term:
$$\frac{{\color{red}{\int{\left(\cosh{\left(2 x \right)} - 1\right)d x}}}}{2} = \frac{{\color{red}{\left(- \int{1 d x} + \int{\cosh{\left(2 x \right)} d x}\right)}}}{2}$$
Apply the constant rule $$$\int c\, dx = c x$$$ with $$$c=1$$$:
$$\frac{\int{\cosh{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{\int{1 d x}}}}{2} = \frac{\int{\cosh{\left(2 x \right)} d x}}{2} - \frac{{\color{red}{x}}}{2}$$
Let $$$u=2 x$$$.
Then $$$du=\left(2 x\right)^{\prime }dx = 2 dx$$$ (steps can be seen »), and we have that $$$dx = \frac{du}{2}$$$.
Thus,
$$- \frac{x}{2} + \frac{{\color{red}{\int{\cosh{\left(2 x \right)} d x}}}}{2} = - \frac{x}{2} + \frac{{\color{red}{\int{\frac{\cosh{\left(u \right)}}{2} d u}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(u \right)} = \cosh{\left(u \right)}$$$:
$$- \frac{x}{2} + \frac{{\color{red}{\int{\frac{\cosh{\left(u \right)}}{2} d u}}}}{2} = - \frac{x}{2} + \frac{{\color{red}{\left(\frac{\int{\cosh{\left(u \right)} d u}}{2}\right)}}}{2}$$
The integral of the hyperbolic cosine is $$$\int{\cosh{\left(u \right)} d u} = \sinh{\left(u \right)}$$$:
$$- \frac{x}{2} + \frac{{\color{red}{\int{\cosh{\left(u \right)} d u}}}}{4} = - \frac{x}{2} + \frac{{\color{red}{\sinh{\left(u \right)}}}}{4}$$
Recall that $$$u=2 x$$$:
$$- \frac{x}{2} + \frac{\sinh{\left({\color{red}{u}} \right)}}{4} = - \frac{x}{2} + \frac{\sinh{\left({\color{red}{\left(2 x\right)}} \right)}}{4}$$
Therefore,
$$\int{\sinh^{2}{\left(x \right)} d x} = - \frac{x}{2} + \frac{\sinh{\left(2 x \right)}}{4}$$
Add the constant of integration:
$$\int{\sinh^{2}{\left(x \right)} d x} = - \frac{x}{2} + \frac{\sinh{\left(2 x \right)}}{4}+C$$
Answer
$$$\int \sinh^{2}{\left(x \right)}\, dx = \left(- \frac{x}{2} + \frac{\sinh{\left(2 x \right)}}{4}\right) + C$$$A