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Question Number 42501 by maxmathsup by imad last updated on 26/Aug/18

$$calculateI={\int }_0^{2\pi }\frac{cos\left(4x\right)}{cosx+sinx}andJ={\int }_0^{2\pi }\frac{sin\left(4x\right)}{cosx+sinx}dx$$

Commented by maxmathsup by imad last updated on 27/Aug/18

$$wehaveI={\int }_0^{\pi }\frac{cos\left(4x\right)}{cosx+sinx}dx+{\int }_{\pi }^{2\pi }\frac{cos\left(4x\right)}{cosx+sinx}butchangement$$$$x=\pi +tgive{\int }_{\pi }^{2\pi }\frac{cos\left(4x\right)}{cosx+sinx}dx={\int }_0^{\pi }\frac{cos\left(4t\right)}{-cost-sint}dt\Rightarrow I=0$$$$thesamemethodshowthatJ=0$$

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