Metamath Proof Explorer


Theorem itgsin0pi

Description: Calculation of the integral for sine on the (0,π) interval. (Contributed by Glauco Siliprandi, 29-Jun-2017)

Ref Expression
Assertion itgsin0pi 0 π sin x dx = 2

Proof

Step Hyp Ref Expression
1 eqid t 0 π cos t = t 0 π cos t
2 1 itgsin0pilem1 0 π sin x dx = 2