Question

Ajutor, vă rog frumos!

Answer 1

$\displaystyle \int_{-\frac{1}{2}}^{\frac{1}{2}}\text{sign } x\, \sin \dfrac{\pi x}{2}\, dx =\\ \\ = \int_{-\frac{1}{2}}^{0}\text{sign } x\, \sin \dfrac{\pi x}{2}\, dx+ \int_{0}^{\frac{1}{2}}\text{sign }x\, \sin \dfrac{\pi x}{2}\, dx = \\ \\ = \int_{-\frac{1}{2}}^{0}(-1)\cdot \sin \dfrac{\pi x}{2}\, dx+ \int_{0}^{\frac{1}{2}} 1\cdot \sin \dfrac{\pi x}{2}\, dx =$

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

$=\boxed{\dfrac{4-2\sqrt 2}{\pi}}$