# Math Help - Trig Substitution Problem

1. ## Trig Substitution Problem

$
\int_0^{7 \sqrt{3}/2} \sqrt{49 - x^2} \, dx
$

This problem really made me depressed..

Heres what I did:
$
let x=7sin\theta, dx=7cos\theta, \theta=\arcsin(\frac{x}{7})
$

$
\int\sqrt{49-49\sin^2\theta}dx = 7\int\sqrt{1-\sin^2\theta}dx=7\int\cos\theta7\cos\theta d\theta=49\int\cos^2\theta d\theta
$

Then, I apply the half hangle formula..
$\frac{49}{2}\int1+\cos(2\theta}) d\theta$

This integrates to..

$
\frac{49}{2}(\theta+\frac{\sin(2\theta)}{2})
$

Now, got rid of the 2theta with another crazy identity..

$
\frac{49}{2}(\theta+\sin\theta\cos\theta)
$

Drew an actual triangle, and converted theta to x:

$
\frac{49}{2}(\arcsin(\frac{x}{2})+\frac{x}{7}*\fra c{\sqrt{49-x^2}}{7})
$

Nowww, after all this work.. when I plug in the upper limit of integration it is undefined! Apparently it isnt in the natural domain of arcsin..

How would I go about solving this? Did I integrate this correctly? I considered using hyperbolic substitution (sinh). Would this work? Would all the identities I just used with sin work with sinh aswell? If so, how do I convert theta -> x using hyperbolic trig functions?

Thanks!

2. The anti-derivative should be

$\dfrac{1}{2}\,x\sqrt{49-x^{2}}+\dfrac{49}{2}\,\sin^{-1}\left(\dfrac{x}{7}\right).$

So the argument of your arcsin function is incorrect. I think that might have been a simple oversight. That should correct your problem, because the integrand is perfectly well-defined on the entire region of integration.