# Calculus Integral

• March 8th 2009, 06:42 PM
HMV
Calculus Integral
Int(x/sqr(x^2+4x+8))dx
• March 8th 2009, 06:50 PM
Chris L T521
Quote:

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?
• March 8th 2009, 07:13 PM
HMV
Is int (secx) the right answer?
• March 8th 2009, 07:30 PM
Chris L T521
Quote:

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?
• March 8th 2009, 07:39 PM
HMV
I can certainly work the integral from there. Would you mind explainimg how you got to that simplyfied form of the integral?
• March 8th 2009, 07:55 PM
Chris L T521
Quote:

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?
• March 8th 2009, 08:12 PM
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