# Math Help - Calculus Integral

1. ## Calculus Integral

Int(x/sqr(x^2+4x+8))dx

2. Originally Posted by HMV
Int(x/sqr(x^2+4x+8))dx
This is the same as $\int\frac{x\,dx}{\sqrt{x^2+4x+4+4}}=\int\frac{x\,d x}{\sqrt{(x+2)^2+4}}=\int\frac{x\,dx}{\sqrt{(x+2)^ 2+2^2}}$

Now apply the trigonometric substitution $x+2=2\tan\theta$

Can you try to take it from here?

3. Is int (secx) the right answer?

4. Originally Posted by HMV
Is int (secx) the right answer?
Not quite...

You should simplify the integral down to $2\int \left(\sec\theta\tan\theta-\sec \theta\right)\,d\theta$ after applying the substitution. Keep in mind that $\int \sec\theta\,d\theta=\ln\left|\sec\theta+\tan\theta \right|+C$

Can you continue on?

5. I can certainly work the integral from there. Would you mind explainimg how you got to that simplyfied form of the integral?

6. Originally Posted by HMV
I can certainly work the integral from there. Would you mind explainimg how you got to that simplyfied form of the integral?
Sure

Our integral was $\int\frac{x\,dx}{\sqrt{(x+2)^2+2^2}}$

Applying the trig substitution $x+2=2\tan\theta\implies x=2\left(\tan\theta-1\right)$, we see that $\,dx=2\sec^2\theta\,d\theta$.

Substituting all this into the integral, we have $\int\frac{2\left(\tan\theta-1\right)2\sec^2\theta\,d\theta}{\sqrt{4\left(\tan^ 2\theta+1\right)}}=4\int\frac{\left(\tan\theta-1\right)\sec^2\theta\,d\theta}{2\sec\theta}$ $=2\int\left(\tan\theta-1\right)\sec\theta\,d\theta=2\int\left(\sec\theta\ tan\theta-\sec\theta\right)\,d\theta$

Does this make sense?

7. Here's an alternate way of solving this problem. It might be easier or harder for you, depending on what you're used to.

Setup the integral as $\int {\frac{{x dx}}{{\sqrt {{{(x + 2)}^2} + 4} }}}
$

Let $x + 2 = 2\sinh \theta$, since ${\sinh ^2}\theta + 1 = {\cosh ^2}\theta$

so $dx = 2\cosh \theta d\theta$ and $x = 2\sinh \theta - 2
$

$\int {\frac{{(2\sinh \theta - 2)(2\cosh \theta )d\theta }}{{\sqrt {4{{\sinh }^2}\theta + 4} }}} = \frac{4}{2}\int {\frac{{\sinh \theta \cosh \theta - {{\cosh }}\theta }}{{\sqrt {{{\sinh }^2}\theta + 1} }} d\theta }$

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

$= 2\cosh \theta - 2\theta + c$

remember that $x + 2 = 2\sinh \theta$, so $\theta = {\sinh ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right)
$

use the identity ${\sinh ^2}\theta + 1 = {\cosh ^2}\theta
$
to show:

$\cosh \theta = \sqrt {1 + {{\sinh }^2}\theta } = \sqrt {1 + {{\left( {\frac{{x + 2}}{2}} \right)}^2}} = \sqrt {\frac{{{x^2}}}{4} + x + 2} = \frac{1}{2}\sqrt {{x^2} + 4x + 8}
$

$2\cosh \theta - 2\theta + c = \sqrt {{x^2} + 4x + 8} - 2{\sinh ^{ - 1}}\left( {\frac{{x + 2}}{2}} \right) + c