Question

integrala de la pi/6 la pi/2 din sin2x/sinx dx ?

Answer 1

[tex]\it \dfrac{\sin2x}{ \sin x} =\dfrac{2\sin x \cos x}{sinx} = 2\cos x \\\;\\ \\\;\\ I = \int_{\frac{\pi}{6}} ^{\frac{\pi}{2}} 2\cos x \ dx = 2\sin x \Big{|}_{\frac{\pi}{6}} ^{\frac{\pi}{2}} = 2\sin \dfrac{\pi}{2} -2\sin \dfrac{\pi}{6} = 2\cdot1-2\cdot\dfrac{1}{2} = 1[/tex]