Μια φορά και έναν καιρό ήταν η τριγωνομετρία 2
by mathxl, Apr 7, 2013, 2:48 PM
Όμοια το όλοκλήρωμα ![\[I = \int\limits_0^{\pi /2} {\frac{{1 + sinx}}{{1 + cosx}}} dx\]](//latex.artofproblemsolving.com/7/3/e/73ebc92d6fd9ecc8935ce8a7cfe0f55262db4fd5.png)
θα υπολογιζόταν εύκολα πριν 2 χρόνια....ως εξής
![\[I = \int\limits_0^{\pi /2} {\frac{{1 + sinx}}{{1 + cosx}}} dx = \int\limits_0^{\pi /2} {\frac{1}{{1 + cosx}}} dx + \int\limits_0^{\pi /2} {\frac{{\sin x}}{{1 + cosx}}} dx = \]](//latex.artofproblemsolving.com/1/c/9/1c954b7f8ace079e5fe5d5716429c329864202c8.png)
![\[I = \int\limits_0^{\pi /2} {\frac{{1 + sinx}}{{1 + cosx}}} dx\]](http://latex.artofproblemsolving.com/7/3/e/73ebc92d6fd9ecc8935ce8a7cfe0f55262db4fd5.png)
θα υπολογιζόταν εύκολα πριν 2 χρόνια....ως εξής![\[I = \int\limits_0^{\pi /2} {\frac{{1 + sinx}}{{1 + cosx}}} dx = \int\limits_0^{\pi /2} {\frac{1}{{1 + cosx}}} dx + \int\limits_0^{\pi /2} {\frac{{\sin x}}{{1 + cosx}}} dx = \]](http://latex.artofproblemsolving.com/1/c/9/1c954b7f8ace079e5fe5d5716429c329864202c8.png)
This post has been edited 1 time. Last edited by mathxl, Apr 7, 2013, 2:53 PM
November 2017
February 2014