# What method has to be used to integrate this? Seems to be non-integrable, but is.

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.

• Why not factor out the $5$s immediately? – David G. Stork Apr 8 '19 at 21:13

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.

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.