Originally Posted by

**joatmon** I've gone over this one but can't see where I am going wrong:

$\displaystyle \displaystyle \int_0^2 x^3\sqrt{x^2 + 4}dx$

Here's my work:

let $\displaystyle x = 2 \tan\theta \rightarrow dx = 2 \sec^2\theta$

$\displaystyle \int_0^\frac{\pi}{4} 8 \tan^3\theta2\sec\theta2\sec^2\theta d\theta$

$\displaystyle 32 \int_0^\frac{\pi}{4} (1-\sec^2\theta)\sec^2\thetasec\thetatan\theta d\theta$

$\displaystyle let u = \sec\theta \rightarrow du = \sec\theta\tan\theta d\theta$

$\displaystyle 32 \int_1^{\sqrt{2}} (u^2 - u^4)du$

$\displaystyle \frac{32u^3}{3} - \frac{32u^5}{5}\displaystyle\mid_1^\sqrt{2}$

$\displaystyle \frac{32 \cdot 2^\frac{3}{2}} {3} - \frac{32 \cdot 2^\frac{5}{2}} {5}-\frac{32}{3} + \frac{32}{5}$

When I plug the original problem into my calculator, I get 10.300645 as an answer. I get 1.767311 on my final answer, which is why I don't think that I am getting to the right place.

Can anyone help? Thanks.