Question

integral_0^pi x*sinx/(3+cos^2(x))dx

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Answer 1

$\displaystyle \int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\,dx =\int_{0}^{\pi}\dfrac{(0+\pi - x)\sin(0+\pi-x)}{3+\cos^2(0+\pi-x)}\, dx = \\ \\ =\int_{0}^{\pi}\dfrac{(\pi-x)\sin x}{3+\cos^2 x}\, dx = \int_{0}^{\pi}\dfrac{\pi\sin x}{3+\cos^2 x}\, dx -\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx \\ \\ \\ \\ \Rightarrow \int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx+\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx = \int_{0}^{\pi}\dfrac{\pi \sin x}{3+\cos^2 x}\, dx$

$\displaystyle 2\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx= -\pi\int_{0}^{\pi}\dfrac{(\cos x)'}{\cos^2 x+(\sqrt 3)^2}\, dx\\ \\ \\ 2\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx= -\dfrac{\pi}{\sqrt 3}\arctan\Big(\dfrac{\cos x}{\sqrt 3}\Big)\Big|_{0}^\pi\\ \\ 2\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx = \dfrac{\pi}{\sqrt 3}\arctan\Big(\dfrac{1}{\sqrt 3}\Big)+\dfrac{\pi}{\sqrt 3}\arctan\Big(\dfrac{1}{\sqrt 3}\Big)$

$\displaystyle 2\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx= \dfrac{\pi}{\sqrt 3}\cdot \dfrac{\pi}{6}+ \dfrac{\pi}{\sqrt 3}\cdot \dfrac{\pi}{6}\\ \\ 2\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx = \dfrac{2\pi^2}{6\sqrt 3}\\ \\ \Rightarrow \boxed{\int_{0}^{\pi}\dfrac{x\sin x}{3+\cos^2 x}\, dx= \dfrac{\pi^2}{6\sqrt 3}}$

$\text{M-am folosit de formula lui King:}\\ \\ \displaystyle \int_{a}^bf(x)\, dx = \int_{a}^bf(a+b-x)\, dx$