# How do you integrate int 1/sqrt(x^2-9x+8)  using trigonometric substitution?

Apr 22, 2018

$I = \int \frac{1}{\sqrt{{x}^{2} - 9 x + 8}} \mathrm{dx} = \int \frac{1}{\sqrt{{\left(x - \frac{9}{2}\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}}} \mathrm{dx}$
We know that, $\int \frac{1}{\sqrt{{X}^{2} - {A}^{2}}} \mathrm{dX} = \ln | X + \sqrt{{X}^{2} - {A}^{2}} | + c$
Put,$X = x - \frac{9}{2} , \mathmr{and} A = \frac{7}{2}$,then simplify,term in sq.rt.
$I = \ln | \left(x - \frac{9}{2}\right) + \sqrt{{x}^{2} - 9 x + 8} | + C$

#### Explanation:

With Trig. Subst. answer, see below.

Decide yourself ,which method is better.

We have, x^2-9x+color(red)8=x^2-9x+color(red)(32/4)=x^2- 9x+color(red)(81/4-49/4)

$\therefore {x}^{2} - 9 x + 8 = {\left(x - \frac{9}{2}\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}$

$i . e . I = \int \frac{1}{\sqrt{{\left(x - \frac{9}{2}\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}}} \mathrm{dx}$

Let, $x - \frac{9}{2} = \frac{7}{2} \sec u \implies \mathrm{dx} = \frac{7}{2} \sec u \tan u \mathrm{du}$

So,

$I = \int \frac{\frac{7}{2} \sec u \tan u}{\sqrt{{\left(\frac{7}{2} \sec u\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}}} \mathrm{du}$

$= \int \frac{\frac{7}{2} \sec u \tan u}{\frac{7}{2} \sqrt{{\tan}^{2} u}} \mathrm{du}$

$= \int \frac{\sec u \tan u}{\tan} u \mathrm{du}$

$= \int \sec u \mathrm{du}$

$= \ln | \sec u + \tan u | + c$

$= \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | + c , \to \left[w h e r e , \sec u = \frac{x - \frac{9}{2}}{\frac{7}{2}}\right]$

$= \ln | \frac{x - \frac{9}{2}}{\frac{7}{2}} + \sqrt{{\left(\frac{x - \frac{9}{2}}{\frac{7}{2}}\right)}^{2} - 1} | + c$

$= \ln | \frac{x - \frac{9}{2}}{\frac{7}{2}} + \frac{\sqrt{{\left(x - \frac{9}{2}\right)}^{2} - {\left(\frac{7}{2}\right)}^{2}}}{\frac{7}{2}} | + c$

$= \ln | \frac{x - \frac{9}{2}}{\frac{7}{2}} + \frac{\sqrt{{x}^{2} - 9 x + 8}}{\frac{7}{2}} | + c$

$= \ln | \frac{x - \frac{9}{2} + \sqrt{{x}^{2} - 9 x + 8}}{\frac{7}{2}} | + c$

$= \ln | x - \frac{9}{2} + \sqrt{{x}^{2} - 9 x + 8} | - \ln \left(\frac{7}{2}\right) + c$

$= \ln | x - \frac{9}{2} + \sqrt{{x}^{2} - 9 x + 8} | + C , w h e r e , C = c - \ln \left(\frac{7}{2}\right)$

Apr 22, 2018

$\implies {\cosh}^{- 1} \left(\frac{2 x - 9}{7}\right) + c$

#### Explanation:

=> int 1/ ( sqrt(( x-9/2)^2 -49/4 )

$\implies u = \frac{2 x - 9}{7}$

$\implies \mathrm{dx} = \frac{7}{2} \mathrm{du}$

$\implies \int \frac{7}{\sqrt{49 {u}^{2} - 49}} \mathrm{dx}$

$\implies \int \frac{1}{\sqrt{{u}^{2} - 1}} \mathrm{du}$

$\implies u = \cosh \theta$

$\implies \mathrm{du} = \sinh \theta d \theta$

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

$\implies \int d \theta$

$\implies \theta + c$

$\implies {\cosh}^{- 1} u + c$

$\implies {\cosh}^{- 1} \left(\frac{2 x - 9}{7}\right) + c$