1. ## Trig Integral

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

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

Here's my work:

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

$\int_0^\frac{\pi}{4} 8 \tan^3\theta2\sec\theta2\sec^2\theta d\theta$
$32 \int_0^\frac{\pi}{4} (1-\sec^2\theta)\sec^2\thetasec\thetatan\theta d\theta$

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

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

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

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

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.

2. ## Re: Trig Integral

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

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

Here's my work:

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

$\int_0^\frac{\pi}{4} 8 \tan^3\theta2\sec\theta2\sec^2\theta d\theta$
$32 \int_0^\frac{\pi}{4} (1-\sec^2\theta)\sec^2\thetasec\thetatan\theta d\theta$

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

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

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

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

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.
How did you get $\displaystyle 8\tan^3{\theta}\cdot \sec{\theta} \cdot 2\sec^2{\theta}$ to $\displaystyle 32(1 - \sec^2{\theta})\sec^2{\theta}$?

Surely it would be $\displaystyle 32\tan{\theta}(1 - \sec^2{\theta})\sec^3{\theta}$...

3. ## Re: Trig Integral

Yes, you're right. I left off a $\sec\theta\tan\theta$ off of that line, which becomes my $du$ after the next substitution. I think that this is a typo and doesn't affect the rest of the calculation. Do you see it otherwise?

4. ## Re: Trig Integral

The mistake is that you use $\tan^2\theta =1-\sec^2 \theta$ (which is incorrect) instead of $\tan^2\theta=\sec^2\theta-1$.

5. ## Re: Trig Integral

Originally Posted by joatmon
Yes, you're right. I left off a $\sec\theta\tan\theta$ off of that line, which becomes my $du$ after the next substitution. I think that this is a typo and doesn't affect the rest of the calculation. Do you see it otherwise?