Question

Am nevoie de ajutor pentru integrala aceasta de la UTCN!!!

Answer 1

Aplic formula lui King:

$\displaystyle \int_{a}^{b}f(x)\, dx =\int_{a}^{b}f(a+b-x)\, dx$

$\displaystyle I =\int_{0}^{2\pi}\dfrac{x\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx = \\ \\ =\int_{0}^{2\pi}\dfrac{(0+2\pi - x)\sin^{100}(0+2\pi -x)}{\sin^{100}(0+2\pi -x)+\cos^{100}(0+2\pi -x)}\, dx = \\ \\ = \int_{0}^{2\pi}\dfrac{(2\pi-x)\sin^{100}x}{\sin^{100} x+\cos^{100}x}\, dx = 2\pi\int_{0}^{2\pi}\dfrac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx-I$

$\displaystyle 2I = 2\pi\int_{0}^{2\pi}\dfrac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx\\ \\ \\\text{Teorema: }\\ \text{Fie }T\text{ perioada lui } f(x)\\ \text{Daca }f(T-x) = f(x) \Rightarrow \int_{0}^{T} f(x)\, dx = 2\int_{0}^{\frac{T}{2}}f(x)\, dx\\ \\\\ 2I = 2\pi \cdot 2\int_{0}^{\pi}\dfrac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx \\\\ 2I = 4\pi\cdot 2\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx$

$\displaystyle I = 4\pi\int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx\\ \\ I = 4\pi \int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{100}(\frac{\pi}{2}-x)}{\sin^{100}(\frac{\pi}{2}-x)+\cos^{100}(\frac{\pi}{2}-x)}\, dx \\ \\ I = 4\pi \int_{0}^{\frac{\pi}{2}}\dfrac{\cos^{100}x}{\sin^{100}x+\cos^{100}x}\, dx \\ \\\\ 2I = 4\pi \int_{0}^{\frac{\pi}{2}}\dfrac{\sin^{100}x}{\sin^{100}x+\cos^{100}x}\, dx +4\pi\int_{0}^{\frac{\pi}{2}}\dfrac{\cos^{100}x}{\sin^{100}x+\cos^{100}x}\, dx$

$\displaystyle 2I = 4\pi \int_{0}^{\frac{\pi}{2}} 1\, dx \\ \\ I = 2\pi x\Big|_{0}^{\frac{\pi}{2}} \\ \\ \\\Rightarrow \boxed{I = \pi^2}$