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

Jul 27, 2016

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} l n \left(\sqrt{\frac{{\left(2 x - 3\right)}^{2}}{5} - 1} + \frac{2 x - 3}{\sqrt{5}}\right) + C$

#### Explanation:

int 1/sqrt(4x^2-12x+4) *d x=?

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \int \frac{1}{\sqrt{4 \left({x}^{2} - 3 x + 1\right)}} \cdot d x =$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} \int \frac{1}{\sqrt{{x}^{2} - 3 x + 1}} \cdot d x =$

${x}^{2} - 3 x + 1 = {\left(x - \frac{3}{2}\right)}^{2} - \frac{5}{4} \text{ (complete the square)}$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} \int \frac{1}{\sqrt{{\left(x - \frac{3}{2}\right)}^{2} - \frac{5}{4}}} \cdot d x$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} / 4 - \frac{5}{4}}} \cdot d x$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \cancel{2} \cdot \frac{1}{\cancel{2}} \int \frac{1}{\sqrt{{\left(2 x - 3\right)}^{2} - 5}} \cdot d x$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \int \frac{1}{\sqrt{5 {\left(\frac{2 x - 3}{\sqrt{5}}\right)}^{2} - 1}} \cdot d x$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \int \frac{\frac{\sqrt{5}}{5}}{\sqrt{{\left(\frac{2 x - 3}{\sqrt{5}}\right)}^{2} - 1}} \cdot d x$

$u = \frac{2 x - 3}{\sqrt{5}} \text{ ; } d u = \frac{2 \cdot \sqrt{5}}{5} \cdot d x$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} \int \frac{2 \cdot \frac{\sqrt{5}}{5} \cdot d x}{\sqrt{{\left(\frac{2 x - 3}{\sqrt{5}}\right)}^{2} - 1}}$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} \int \frac{d u}{\sqrt{{u}^{2} - 1}}$

$\int \frac{d u}{\sqrt{{u}^{2} - 1}} = l n \sqrt{{u}^{2} - 1} + u$

$\text{undo substitution}$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} l n \left(\sqrt{{\left(\frac{2 x - 3}{\sqrt{5}}\right)}^{2} - 1} + \frac{2 x - 3}{\sqrt{5}}\right)$

$\int \frac{1}{\sqrt{4 {x}^{2} - 12 x + 4}} \cdot d x = \frac{1}{2} l n \left(\sqrt{\frac{{\left(2 x - 3\right)}^{2}}{5} - 1} + \frac{2 x - 3}{\sqrt{5}}\right) + C$