Integral of $$$\sqrt{x^{2} - 25}$$$
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Find $$$\int \sqrt{x^{2} - 25}\, dx$$$.
Solution
Let $$$x=5 \cosh{\left(u \right)}$$$.
Then $$$dx=\left(5 \cosh{\left(u \right)}\right)^{\prime }du = 5 \sinh{\left(u \right)} du$$$ (steps can be seen »).
Also, it follows that $$$u=\operatorname{acosh}{\left(\frac{x}{5} \right)}$$$.
So,
$$$\sqrt{x^{2} - 25} = \sqrt{25 \cosh^{2}{\left( u \right)} - 25}$$$
Use the identity $$$\cosh^{2}{\left( u \right)} - 1 = \sinh^{2}{\left( u \right)}$$$:
$$$\sqrt{25 \cosh^{2}{\left( u \right)} - 25}=5 \sqrt{\cosh^{2}{\left( u \right)} - 1}=5 \sqrt{\sinh^{2}{\left( u \right)}}$$$
Assuming that $$$\sinh{\left( u \right)} \ge 0$$$, we obtain the following:
$$$5 \sqrt{\sinh^{2}{\left( u \right)}} = 5 \sinh{\left( u \right)}$$$
Integral becomes
$${\color{red}{\int{\sqrt{x^{2} - 25} d x}}} = {\color{red}{\int{25 \sinh^{2}{\left(u \right)} d u}}}$$
Apply the constant multiple rule $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$ with $$$c=25$$$ and $$$f{\left(u \right)} = \sinh^{2}{\left(u \right)}$$$:
$${\color{red}{\int{25 \sinh^{2}{\left(u \right)} d u}}} = {\color{red}{\left(25 \int{\sinh^{2}{\left(u \right)} d u}\right)}}$$
Apply the power reducing formula $$$\sinh^{2}{\left(\alpha \right)} = \frac{\cosh{\left(2 \alpha \right)}}{2} - \frac{1}{2}$$$ with $$$\alpha= u $$$:
$$25 {\color{red}{\int{\sinh^{2}{\left(u \right)} d u}}} = 25 {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}}$$
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 $$$f{\left(u \right)} = \cosh{\left(2 u \right)} - 1$$$:
$$25 {\color{red}{\int{\left(\frac{\cosh{\left(2 u \right)}}{2} - \frac{1}{2}\right)d u}}} = 25 {\color{red}{\left(\frac{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}{2}\right)}}$$
Integrate term by term:
$$\frac{25 {\color{red}{\int{\left(\cosh{\left(2 u \right)} - 1\right)d u}}}}{2} = \frac{25 {\color{red}{\left(- \int{1 d u} + \int{\cosh{\left(2 u \right)} d u}\right)}}}{2}$$
Apply the constant rule $$$\int c\, du = c u$$$ with $$$c=1$$$:
$$\frac{25 \int{\cosh{\left(2 u \right)} d u}}{2} - \frac{25 {\color{red}{\int{1 d u}}}}{2} = \frac{25 \int{\cosh{\left(2 u \right)} d u}}{2} - \frac{25 {\color{red}{u}}}{2}$$
Let $$$v=2 u$$$.
Then $$$dv=\left(2 u\right)^{\prime }du = 2 du$$$ (steps can be seen »), and we have that $$$du = \frac{dv}{2}$$$.
So,
$$- \frac{25 u}{2} + \frac{25 {\color{red}{\int{\cosh{\left(2 u \right)} d u}}}}{2} = - \frac{25 u}{2} + \frac{25 {\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2}$$
Apply the constant multiple rule $$$\int c f{\left(v \right)}\, dv = c \int f{\left(v \right)}\, dv$$$ with $$$c=\frac{1}{2}$$$ and $$$f{\left(v \right)} = \cosh{\left(v \right)}$$$:
$$- \frac{25 u}{2} + \frac{25 {\color{red}{\int{\frac{\cosh{\left(v \right)}}{2} d v}}}}{2} = - \frac{25 u}{2} + \frac{25 {\color{red}{\left(\frac{\int{\cosh{\left(v \right)} d v}}{2}\right)}}}{2}$$
The integral of the hyperbolic cosine is $$$\int{\cosh{\left(v \right)} d v} = \sinh{\left(v \right)}$$$:
$$- \frac{25 u}{2} + \frac{25 {\color{red}{\int{\cosh{\left(v \right)} d v}}}}{4} = - \frac{25 u}{2} + \frac{25 {\color{red}{\sinh{\left(v \right)}}}}{4}$$
Recall that $$$v=2 u$$$:
$$- \frac{25 u}{2} + \frac{25 \sinh{\left({\color{red}{v}} \right)}}{4} = - \frac{25 u}{2} + \frac{25 \sinh{\left({\color{red}{\left(2 u\right)}} \right)}}{4}$$
Recall that $$$u=\operatorname{acosh}{\left(\frac{x}{5} \right)}$$$:
$$\frac{25 \sinh{\left(2 {\color{red}{u}} \right)}}{4} - \frac{25 {\color{red}{u}}}{2} = \frac{25 \sinh{\left(2 {\color{red}{\operatorname{acosh}{\left(\frac{x}{5} \right)}}} \right)}}{4} - \frac{25 {\color{red}{\operatorname{acosh}{\left(\frac{x}{5} \right)}}}}{2}$$
Therefore,
$$\int{\sqrt{x^{2} - 25} d x} = \frac{25 \sinh{\left(2 \operatorname{acosh}{\left(\frac{x}{5} \right)} \right)}}{4} - \frac{25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}$$
Using the formulas $$$\sin{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\sin{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{1 - \alpha^{2}}$$$, $$$\cos{\left(2 \operatorname{asin}{\left(\alpha \right)} \right)} = 1 - 2 \alpha^{2}$$$, $$$\cos{\left(2 \operatorname{acos}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, $$$\sinh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha^{2} + 1}$$$, $$$\sinh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha \sqrt{\alpha - 1} \sqrt{\alpha + 1}$$$, $$$\cosh{\left(2 \operatorname{asinh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} + 1$$$, $$$\cosh{\left(2 \operatorname{acosh}{\left(\alpha \right)} \right)} = 2 \alpha^{2} - 1$$$, simplify the expression:
$$\int{\sqrt{x^{2} - 25} d x} = \frac{5 x \sqrt{\frac{x}{5} - 1} \sqrt{\frac{x}{5} + 1}}{2} - \frac{25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}$$
Simplify further:
$$\int{\sqrt{x^{2} - 25} d x} = \frac{x \sqrt{x - 5} \sqrt{x + 5} - 25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}$$
Add the constant of integration:
$$\int{\sqrt{x^{2} - 25} d x} = \frac{x \sqrt{x - 5} \sqrt{x + 5} - 25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2}+C$$
Answer
$$$\int \sqrt{x^{2} - 25}\, dx = \frac{x \sqrt{x - 5} \sqrt{x + 5} - 25 \operatorname{acosh}{\left(\frac{x}{5} \right)}}{2} + C$$$A