# How do you integrate int 1/sqrt(4x^2-12x+10)  using trigonometric substitution?

Mar 4, 2018

$\implies \frac{1}{2} {\sinh}^{- 1} \left(2 x - 3\right) + c$

#### Explanation:

The first thing to do is rewrite $4 {x}^{2} - 12 x + 10$ :

$4 {x}^{2} - 12 x + 10 \equiv 4 {\left(x - \frac{3}{2}\right)}^{2} + 1$

$\implies \int \frac{\mathrm{dx}}{\sqrt{4 {\left(x - \frac{3}{2}\right)}^{2} + 1}}$

Making the substitution:

$2 \left(x - \frac{3}{2}\right) = \sinh \theta$

$2 \mathrm{dx} = \cosh \theta d \theta$

$4 {\left(x - \frac{3}{2}\right)}^{2} = {\sinh}^{2} \theta$

$\implies \int \frac{\frac{1}{2} \cosh \theta d \theta}{\sqrt{{\sinh}^{2} \theta + 1}}$

We know ${\cosh}^{2} \theta - {\sinh}^{2} \theta = 1 \implies 1 + {\sinh}^{2} = {\cosh}^{2} \theta$

$\implies \frac{1}{2} \int \frac{\cosh \theta d \theta}{\cosh} \theta$

$\implies \frac{1}{2} \int 1 d \theta$

$\implies \frac{1}{2} \theta + c$

$2 x - 3 = \sinh \theta \implies \theta = {\sinh}^{- 1} \left(2 x - 3\right)$

$\implies \frac{1}{2} {\sinh}^{- 1} \left(2 x - 3\right) + c$