Verifying trigonometric integral result

Question

I'm trying to calculate the below integral:

$$ \int_{0}^{2 \pi} \sqrt{(-5\sin t-10\cos t\sin t)^2+(-5\sin^2t+5\cos^2t+5\cos t)^2} dt $$

After some lengthy manipulations, I get the below as the result of the indefinite integral:

$$ 20\sin\frac{t}{2} $$

This seems right as confirmed by Wolframalpha and several other online integral calculators. Now, the answer sheet says that the result is $40$ (again: confirmed by Wolframalpha), but how could that be - when you plug $2\pi$ and $0$ as the bounds of integration, the result should be $0$. Or did my brain just turn off?

Answer 1

It's because you need to take into account the modulus sign!

You have $$10\int_{0}^{2\pi}\sqrt{\cos^{2}(\frac{t}{2})}dt=10\int_{0}^{2\pi}|\cos(\frac{t}{2})|dt$$ $$=10\big[\int_{0}^{\pi}\cos(\frac{t}{2})dt-\int_{\pi}^{2\pi}\cos(\frac{t}{2})dt\big]$$ $$=20\big[\sin(\frac{t}{2})\big|_{t=0}^{t=\pi}-\sin(\frac{t}{2})\big|_{t=\pi}^{t=2\pi}\big]$$ $$=20\big[1-0-(0-1)\big]=20\cdot2=40.$$

Answer 2

I obtained

$$ I = \int_{0}^{2\pi} 10 |cos t/2|dt \\ = \int_{0}^{\pi} 20 |cos u|du \\ = 2 \int_{0}^{\frac{\pi}{2}} 20 |cos u|du \\ = 40 $$

Here I have used property of even functions to change limits from $0$ to $\pi/2$. Integrand is zero at endpoints but not in between, so integral won't be zero.