Integration with $\sin(t)$ as the variable.

Question

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 $\sin(2x) = 2\sin x\cos x$?

Just make use of the fact that $d\sin{(t)}=\cos{(t)}dt$, the first integral you wrote down understood in the Riemann Stieltjes sense — shdp, Mar 10 '17 at 00:47
Possibly relevant: https://math.stackexchange.com/questions/295383/integration-of-a-function-with-respect-to-another-function — David K, May 17 '17 at 13:50

score 4 · Accepted Answer · answered Mar 10 '17 at 00:49

4

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?

answered Mar 10 '17 at 00:49

Fimpellizzeri

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1

I got the answer, but where did the $\frac{1}{2}$ in the integral come from in the solution someone else provided? – Felicio Grande Mar 10 '17 at 00:54
Ah I see. Thanks. – Felicio Grande Mar 10 '17 at 00:54

Integration with $\sin(t)$ as the variable.

1 Answers1