# How do you integrate int x/sqrt(3x^2-6x-8) dx using trigonometric substitution?

Jun 28, 2018

intx/sqrt(3x²-6x-8)dx=sqrt11/3cot(csc^(-1)(sqrt(3/11)(x-1)))+1/sqrt3ln(|cot(csc^(-1)(sqrt(3/11)(x-1)))+sqrt(3/11)(x-1)|)+C, $C \in \mathbb{R}$

#### Explanation:

intx/sqrt(3x²-6x-8)dx

=1/sqrt(3)intx/sqrt(x²-2x-8/3)dx

=1/sqrt(3)intx/sqrt((x-1)²-11/3)

Let $x - 1 = \sqrt{\frac{11}{3}} \csc \left(\theta\right)$

$\mathrm{dx} = - \sqrt{\frac{11}{3}} \cot \left(\theta\right) \csc \left(\theta\right) d \theta$

1/sqrt(3)intx/sqrt((x-1)²-11/3)=-1/sqrt(3)int((sqrt(11/3)csc(theta)+1)cot(theta)csc(theta))/sqrt(csc²(theta)-1)d theta
Because csc²(x)-1=cot²(x),
=-1/sqrt(3)(intsqrt(11/3)csc²(theta)d theta + intcsc(theta)d theta)
$= \frac{\sqrt{11}}{3} \cot \left(\theta\right) + \frac{1}{\sqrt{3}} \ln \left(| \cot \left(\theta\right) + \csc \left(\theta\right) |\right)$
Finally, $\theta = {\csc}^{- 1} \left(\sqrt{\frac{3}{11}} \left(x - 1\right)\right)$
So:
intx/sqrt(3x²-6x-8)dx=sqrt11/3cot(csc^(-1)(sqrt(3/11)(x-1)))+1/sqrt3ln(|cot(csc^(-1)(sqrt(3/11)(x-1)))+sqrt(3/11)(x-1)|)+C, $C \in \mathbb{R}$