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I am currently working on a question about Calculus on Manifolds and ran into an integral that I'm confused about. How would you evaluate:

$$\int_0^{2\pi}\cos(t) d\sin(t)$$

The answer is $\pi$, by simplifying the above to: $$\frac{1}{2}\cdot \int_0^{2\pi}dt$$ I tried using u-sub but I think I'm missing something here. Am I supposed to use the identity $\sin(2x) = 2\sin x\cos x$?

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This is the Riemann-Stieltjes integral, a generalization of the Riemann integral. Since $\sin(t)$ is everywhere continuously differentiable, we have that the integral is equivalent to

$$\int_{0}^{2\pi}\cos(t)\left(\frac{d}{dt}\sin(t)\right)\,dt=\int_{0}^{2\pi}{\big(\cos(t)\big)}^2\,dt$$

Do you think you can take it from here?

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    $\begingroup$ I got the answer, but where did the $\frac{1}{2}$ in the integral come from in the solution someone else provided? $\endgroup$ – Felicio Grande Mar 10 '17 at 0:54
  • $\begingroup$ Ah I see. Thanks. $\endgroup$ – Felicio Grande Mar 10 '17 at 0:54

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