Find the integration of :

(x^3)/(sqrt(x^2+100))

I did the following work, but it seems my answer is not matching to the Wolfram|Alpha' answer.

Attachment 15355

int x^3/(sqrt( x^2 +100)) dx - Wolfram|Alpha

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- Feb 11th 2010, 02:56 AMracewithferrariIntegration by Trig substitution
Find the integration of :

(x^3)/(sqrt(x^2+100))

I did the following work, but it seems my answer is not matching to the Wolfram|Alpha' answer.

Attachment 15355

int x^3/(sqrt( x^2 +100)) dx - Wolfram|Alpha - Feb 11th 2010, 03:24 AMdedust
- Feb 11th 2010, 03:32 AMProve It
Don't use a trig sub, use a hyperbolic sub.

Let $\displaystyle x = 10\sinh{t}$ so that $\displaystyle dx = 10\cosh{t}\,dt$.

Then the integral becomes

$\displaystyle \int{\frac{x^3}{\sqrt{x^2 + 100}}\,dx}$

$\displaystyle = \int{\frac{1000\sinh^3{t}}{\sqrt{(10\sinh{t})^2 + 100}}\,10\cosh{t}\,dt}$

$\displaystyle = \int{\frac{1000\sinh^3{t}}{10\cosh{t}}\,10\cosh{t} \,dt}$

$\displaystyle = \int{1000\sinh^3{t}\,dt}$

$\displaystyle = \int{1000\sinh{t}\sinh^2{t}\,dt}$

$\displaystyle = \int{1000\sinh{t}(\cosh^2{t} - 1)\,dt}$

$\displaystyle = 1000\int{\sinh{t}(\cosh^2{t} - 1)\,dt}$

Let $\displaystyle u = \cosh{t}$ so that $\displaystyle du = \sinh{t}\,dt$

The integral becomes

$\displaystyle 1000\int{u^2 - 1\,du}$

$\displaystyle = 1000\left(\frac{1}{3}u^3 - u\right) + C$

Now convert it back to a function of $\displaystyle x$.