# How do you integrate int x^2/sqrt(x^2+1) by trigonometric substitution?

May 26, 2018

$\int \frac{{x}^{2} \cdot \mathrm{dx}}{\sqrt{{x}^{2} + 1}} = \frac{x}{2} \cdot \sqrt{{x}^{2} + 1} - \frac{1}{2} {\sinh}^{-} 1 x + C$

#### Explanation:

$\int \frac{{x}^{2} \cdot \mathrm{dx}}{\sqrt{{x}^{2} + 1}}$

After using $x = \sinh y$ and $\mathrm{dx} = \cosh y \cdot \mathrm{dy}$ transforms, this integral became

$\int \frac{{\left(\sinh y\right)}^{2} \cdot \cosh y \cdot \mathrm{dy}}{\sqrt{{\left(\sinh y\right)}^{2} + 1}}$

=$\int \frac{{\left(\sinh y\right)}^{2} \cdot \cosh y \cdot \mathrm{dy}}{\sqrt{{\left(\cosh y\right)}^{2}}}$

=$\int \frac{{\left(\sinh y\right)}^{2} \cdot \cosh y \cdot \mathrm{dy}}{\cosh} y$

=$\int {\left(\sinh y\right)}^{2} \cdot \mathrm{dy}$

=$\int \frac{\cosh 2 y - 1}{2} \cdot \mathrm{dy}$

=$\frac{1}{4} \sinh 2 y - \frac{y}{2} + C$

=$\frac{1}{4} \cdot 2 \sinh y \cdot \cosh y - \frac{y}{2} + C$

=$\frac{1}{2} \sinh y \cdot \cosh y - \frac{y}{2} + C$

After using $x = \sinh y$, $\cosh y = \sqrt{{x}^{2} + 1}$ and $y = {\sinh}^{-} 1 x$ inverse transforms, I found

$\int \frac{{x}^{2} \cdot \mathrm{dx}}{\sqrt{{x}^{2} + 1}} = \frac{x}{2} \cdot \sqrt{{x}^{2} + 1} - \frac{1}{2} {\sinh}^{-} 1 x + C$