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This is my first question on Math Stack Exchange, so thanks in advance for any help, and hope I type this right.

I was asked as a homework question to (basically) solve for S, where $$S=\int_0^{2\pi} T \,dx$$ and T was something long and complicated that I didn't feel like typing through MathJax lol. Anyway, the indefinite integral itself wasn't too hard, but I didn't get the correct definite answer. So I checked the solution, and the first step of the solution was $$\int_0^{2\pi} T \,dx = 2\int_0^{\pi} T \,dx$$ And I was wondering if that is a valid "move," so to speak, and if so, what is the explicit rule/when can it actually be used? Is the rule as simple as $$\int_{Ax_1}^{Ax_2} T \,dx = A\int_{x_1}^{x_2} T \,dx$$ or are there more constraints for when this can be used? Once again, thanks to everyone!

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  • $\begingroup$ Is $T$ a $2\pi$-periodic and even function? $\endgroup$ – Mark Viola Oct 17 '17 at 3:00
  • $\begingroup$ Or simply $\pi$-periodic? $\endgroup$ – Xander Henderson Oct 17 '17 at 3:02
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The rule is not what you're thinking and depends on the form of the $T(x)$ that you didn't feel like writing out. The function (which I'm guessing is some combination of trig functions) probably has symmetry $T(x) = T(2\pi-x)$ in which case we have $$ \int_0^{2\pi} T(x)\;dx = \int_0^\pi T(x)\;dx + \int_\pi^{2\pi} T(x)\;dx \\ = \int_0^\pi T(x)dx + \int_{0}^\pi T(2\pi-u) \; du \\= \int_0^\pi T(x)\;dx + \int_0^\pi T(x)\;dx \\=2\int_0^\pi T(x)\;dx $$ where in the second line we used the substitution $u = 2\pi-x$ and in the third we used the fact that $T(x) = T(2\pi-x)$ and changed the dummy variable back to $x$.

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  • $\begingroup$ Well, guess I should just include the whole problem next time lol. But yes, I believe it is symmetric. The original integral was $$S=\int_0^{2\pi} \sqrt{(a(1- \cos (x)))^2 + (a \sin (x))^2} \,dx$$ Which I believe simplifies down to $$S=\sqrt{2}a\int_0^{2\pi} \sqrt{1-\cos (x)} \,dx$$ and that's why it would be symmetric. That makes a lot of sense, thanks! While I've got you here though, when would that rule need to be used? Because when I took the definite integral by hand, it just cancels out and is 0 $\endgroup$ – Chris Austin Oct 17 '17 at 5:07

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