# How do you integrate int 1/sqrt(4x^2+16x+8)  using trigonometric substitution?

Jun 16, 2018

$\frac{1}{2} \ln | x + 2 + \sqrt{{x}^{2} + 4 x + 2} | + C$

#### Explanation:

It is in general a good idea to divide any polynomial that we have to work with through by its leading coefficient; this almost always makes it simpler to work with. So:

$\int \frac{1}{\sqrt{4 {x}^{2} + 16 x + 8}} \mathrm{dx} = \frac{1}{2} \int \frac{1}{\sqrt{{x}^{2} + 4 x + 2}} \mathrm{dx}$

Complete the square:

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

Make a substitution to cause the square root to vanish via a trig identity. In this case we'll use $x + 2 = \sqrt{2} \sec \theta$. We could also use $\csc$ (which would be similar) or, straying into hyperbolic functions, $\cosh$ (which would be easier).

As $x = - 2 + \sqrt{2} \sec \theta$, $\frac{\mathrm{dx}}{d \theta} = \sqrt{2} \sec \theta \tan \theta$, and so our integral becomes

$\frac{1}{2} \int \frac{1}{\sqrt{2 {\sec}^{2} \theta - 2}} \cdot \sqrt{2} \sec \theta \tan \theta d \theta$

Recalling the identity ${\tan}^{2} \theta = {\sec}^{2} \theta - 1$, we obtain

$\frac{1}{2} \int \frac{1}{\sqrt{2} \tan \theta} \cdot \sqrt{2} \sec \theta \tan \theta d \theta$

which simplifies considerably to become

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

The integral of $\sec x$ is $\ln | \sec x + \tan x |$. Two demonstrations of this are given here: https://www.math.ubc.ca/~feldman/m121/secx.pdf
So we get

$\frac{1}{2} \ln | \sec \theta + \tan \theta | + C$

Transform back to our original variable, $x$:

$\frac{1}{2} \ln | \frac{x + 2}{\sqrt{2}} + \sqrt{{\left(x + 2\right)}^{2} / 2 - 1} | + C$

Rearrange a little to obtain a simpler form:

$\frac{1}{2} \ln | \frac{x + 2}{\sqrt{2}} + \sqrt{\frac{{\left(x + 2\right)}^{2} - 2}{2}} | + C$
$\frac{1}{2} \ln | x + 2 + \sqrt{{\left(x + 2\right)}^{2} - 2} | - \frac{1}{2} \ln \sqrt{2} + C$

Note that the $- \frac{1}{2} \ln \sqrt{2}$ term is a constant and can therefore be folded into the integration constant $C$

$\frac{1}{2} \ln | x + 2 + \sqrt{{x}^{2} + 4 x + 2} | + C$

If we wished to write this in the same style as the original question formula, we could use the same observation about the integration constant to add $\frac{1}{2} \ln 2$ without changing the answer:

$\frac{1}{2} \ln | x + 2 + \sqrt{{x}^{2} + 4 x + 2} | + C$
$\frac{1}{2} \ln | x + 2 + \sqrt{{x}^{2} + 4 x + 2} | + \frac{1}{2} \ln 2 + C$
$\frac{1}{2} \ln | 2 \left(x + 2\right) + \sqrt{4 {x}^{2} + 16 x + 8} | + C$

It's arguable whether it's better to write it so that the expression inside the root matches that in the original integral or so that it is a simpler expression. I think I prefer our first answer myself:

$\frac{1}{2} \ln | x + 2 + \sqrt{{x}^{2} + 4 x + 2} | + C$

It's more elegant.

Jun 16, 2018

$I = \frac{1}{2} \ln | \left(x + 2\right) + \sqrt{{x}^{2} + 4 x + 2} | + C$

#### Explanation:

Here,

$I = \int \frac{1}{\sqrt{4 {x}^{2} + 16 x + 8}} \mathrm{dx}$

$I = \int \frac{1}{2 \sqrt{{x}^{2} + 4 x + 2}} \mathrm{dx}$

$I = \frac{1}{2} \int \frac{1}{\sqrt{{x}^{2} + 4 x + 4 - 2}} \mathrm{dx}$

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

Subst. $x + 2 = \sqrt{2} \sec u \implies \mathrm{dx} = \sqrt{2} \sec u \tan u \mathrm{du}$

$\implies \sec u = \frac{x + 2}{\sqrt{2}}$

$I = \frac{1}{2} \int \frac{1}{\sqrt{2 {\sec}^{2} u - 2}} \times \sqrt{2} \sec u \tan u \mathrm{du}$

$\therefore I = \frac{1}{2} \int \frac{\sqrt{2} \sec u \tan u}{\sqrt{2} \tan u} \mathrm{du}$

$\implies = \frac{1}{2} \int \sec u \mathrm{du}$

$\therefore I = \frac{1}{2} \ln | \sec u + \tan u | + c$

$= \frac{1}{2} \ln | \sec u + \sqrt{{\sec}^{2} u - 1} | + c$

Subst. back , $\sec u = \frac{x + 2}{\sqrt{2}}$

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

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

$I = \frac{1}{2} \ln | \frac{\left(x + 2\right) + \sqrt{{x}^{2} + 4 x + 2}}{\sqrt{2}} | + c$

$I = \frac{1}{2} \ln | \left(x + 2\right) + \sqrt{{x}^{2} + 4 x + 2} | - \frac{1}{2} \ln \sqrt{2} + c$

$I = \frac{1}{2} \ln | \left(x + 2\right) + \sqrt{{x}^{2} + 4 x + 2} | + C ,$
 where,C=c-1/2lnsqrt2