# How do you integrate int 1/(x^4sqrt(x^2+3)) by trigonometric substitution?

May 30, 2018

$\frac{\sqrt{{x}^{2} + 3}}{9 x} - {\left({x}^{2} + 3\right)}^{\frac{3}{2}} / \left(27 {x}^{3}\right) + C$

#### Explanation:

$\int \frac{\mathrm{dx}}{{x}^{4} \cdot \sqrt{{x}^{2} + 3}}$

After using $x = \sqrt{3} \cdot \tan y$ and $\mathrm{dx} = \sqrt{3} \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}$ transforms, I found

$\int \frac{\sqrt{3} \cdot {\left(\sec y\right)}^{2} \cdot \mathrm{dy}}{{\left(\sqrt{3} \cdot \tan y\right)}^{4} \cdot \sqrt{{\left(\sqrt{3} \cdot \tan y\right)}^{2} + 3}}$

=int (sqrt3*(secy)^2*dy)/((9(tany)^4*sqrt((sqrt3*secy)^2))

=int (sqrt3*(secy)^2*dy)/((9(tany)^4*sqrt3*secy)

=$\frac{1}{9} \int \frac{\sec y \cdot \mathrm{dy}}{\tan y} ^ 4$

=$\frac{1}{9} \int {\left(\cot y\right)}^{4} \cdot \sec y \cdot \mathrm{dy}$

=$\frac{1}{9} \int {\left(\cos \frac{y}{\sin} y\right)}^{4} \cdot \left(\frac{\mathrm{dy}}{\cos} y\right)$

=$\frac{1}{9} \int \frac{{\left(\cos y\right)}^{3} \cdot \mathrm{dy}}{\sin y} ^ 4$

=$\frac{1}{9} \int \frac{{\left(\cos y\right)}^{2} \cdot \cos y \cdot \mathrm{dy}}{\sin y} ^ 4$

=$\frac{1}{9} \int \frac{\left(1 - {\left(\sin y\right)}^{2}\right) \cdot \cos y \cdot \mathrm{dy}}{\sin y} ^ 4$

After using $z = \sin y$ and $\mathrm{dz} = \cos y \cdot \mathrm{dy}$ transforms, it became

$\frac{1}{9} \int \frac{\left(1 - {z}^{2}\right) \cdot \mathrm{dz}}{z} ^ 4$

=$\frac{1}{9} \int {z}^{- 4} \cdot \mathrm{dz} - \frac{1}{9} \int {z}^{- 2} \cdot \mathrm{dz}$

=$\frac{1}{9} {z}^{- 1} - \frac{1}{27} {z}^{- 3} + C$

=$\frac{1}{9} {\left(\sin y\right)}^{- 1} - \frac{1}{27} {\left(\sin y\right)}^{- 3} + C$

=$\frac{1}{9} \csc y - \frac{1}{27} {\left(\csc y\right)}^{3} + C$

After using $x = \sqrt{3} \cdot \tan y$, $\tan y = \frac{x}{\sqrt{3}}$, $\sec y = \frac{\sqrt{{x}^{2} + 3}}{\sqrt{3}}$ and $\csc y = \sec \frac{y}{\tan} y = \frac{\sqrt{{x}^{2} + 3}}{x}$ inverse transforms, I found

$\frac{\sqrt{{x}^{2} + 3}}{9 x} - {\left({x}^{2} + 3\right)}^{\frac{3}{2}} / \left(27 {x}^{3}\right) + C$