# How do you integrate int 1/sqrt(x^2-16x+3)  using trigonometric substitution?

Mar 14, 2018

Use the substitution $x - 8 = \sqrt{61} \sec \theta$.

#### Explanation:

Let

$I = \int \frac{1}{\sqrt{{x}^{2} - 16 x + 3}} \mathrm{dx}$

Complete the square in the square root:

$I = \int \frac{1}{\sqrt{{\left(x - 8\right)}^{2} - 61}} \mathrm{dx}$

Apply the substitution $x - 8 = \sqrt{61} \sec \theta$:

$I = \int \frac{1}{\sqrt{61} \tan \theta} \left(\sqrt{61} \sec \theta \tan \theta d \theta\right)$

Simplify:

$I = \int \sec \theta d \theta$

Integrate directly:

$I = \ln | \sec \theta + \tan \theta | + C$

Rescale $C$:

$I = \ln | \sqrt{61} \sec \theta + \sqrt{61} \tan \theta | + C$

Reverse the substitution:

$I = \ln | x - 8 + \sqrt{{x}^{2} - 16 x + 3} | + C$

Mar 14, 2018

$I = \log | \left(x - 8\right) + \sqrt{{x}^{2} - 16 x + 3} | + C$, where,
$C = - \log \sqrt{61} + c$.
NOTE : -It is better to use $\int \frac{1}{\sqrt{{X}^{2} + k}} \mathrm{dX} = \ln | x + \sqrt{{x}^{2} + k} | + c$, from (A)

#### Explanation:

I=int1/sqrt(x^2-16x+3)dx=int1/sqrt(x^2-16x+64-61)dx =int1/sqrt((x-8)^2-(sqrt61)^2)dx,.(A)
We take,$x - 8 = \sqrt{61} \sec \theta \implies x = 8 + \sqrt{61} \sec \theta \implies \mathrm{dx} = \sqrt{61} \sec \theta \tan \theta \cdot d \left(\theta\right)$
$I = \int \frac{\left(\sqrt{61} \sec \theta \tan \theta\right)}{\sqrt{61 {\sec}^{2} \theta - 61}} d \left(\theta\right)$
$I = \int \frac{\left(\sqrt{61} \sec \theta \tan \theta\right)}{\sqrt{61} \sqrt{{\sec}^{2} \theta - 1}} d \left(\theta\right)$$= \int \frac{\sec \theta \tan \theta}{\tan \theta} d \left(\theta\right) = \int \sec \theta d \left(\theta\right)$$= \log | \sec \theta + \tan \theta | + c$
$I = \log | \sec \theta + \sqrt{{\sec}^{2} \theta - 1} | + c$
$I = \log | \frac{x - 8}{\sqrt{61}} + \sqrt{{\left(\frac{x - 8}{\sqrt{61}}\right)}^{2} - 1} | + c$
$I = \log | \frac{x - 8}{\sqrt{61}} + \frac{\sqrt{{\left(x - 8\right)}^{2} - 61}}{\sqrt{61}} | + c$
$I = \log | \left(x - 8\right) + \sqrt{{x}^{2} - 16 x + 3} | - \log | \sqrt{61} | + c$