I'm doing an special assignment for my calculus class and I have to integrate the following in order to obtain the distance of a certain epicycloid from $0$ to $2\pi$. I don't believe the specific epicycloid is relevant to the question. Heres the integral: $$4 \int_{0}^{\frac{\pi}{2}} \sqrt{(-5\sin(t)+5\sin(5t))^2 + (5\cos(t)-5\cos(5t))^2} \; \mathrm{d}t $$ Wolfram Alpha tells me the result is $40$, however when I try to apply Barrow to the indefinite integral, I get the result is undefined. This makes sense, since the indefinite integral is $$ \frac{-5 \cos(2 t) \sqrt{\sin^2(2 t)}}{\sin(2 t)}$$ and the sine of $2 \cdot \frac{\pi}{2}$ and the sine of $0$ is, obviously, $0$. So that's undefined, can't be solved. However Wolfram comes up with $40$ and I've got no idea how.
$\begingroup$
$\endgroup$
1
-
$\begingroup$ Why not factor out the $5$s immediately? $\endgroup$ – David G. Stork Apr 8 '19 at 21:13
Add a comment
|
$\begingroup$
$\endgroup$
When in doubt, plot:
The integral is indeed $40$.
Factor the $5$ out of the integrand and note that
$$\sqrt{(- \sin (t) + \sin (5 t))^2 + (\cos (t) - \cos (5 t))^2} = \sin^2 (2 t)$$
(The integral of $\sin^2 (x)$ is straightforward.)
That the denominator vanishes at the endpoints of integration doesn't mean the full result does.
answered Apr 8 '19 at 21:17
$\begingroup$
$\endgroup$
Whenever I see $\sqrt{\sin^2(2t)}$ in a context where $\sin(2t)\ge 0$, I always like to replace it with $\sin(2t)$. This might help!
Having said that, if your indefinite integral is correct, you get the answer $-40$, which doesn't look right.
