Integralas ir trigonometrinė lygtis

Išspręskite lygtį  :  [tex]\int_{0}^{\pi} (\sin x+x\cos x)dx=\sin 2x[/tex]

$$(x\sin x)'=x'\cdot \sin x+(\sin x)'\cdot x=\sin x+x\cos x.\\\int_{0}^{\pi} (\sin x+x\cos x)dx=(x\sin x)\Biggr|_{0}^{\pi}=\pi\sin(\pi)-0\sin 0=0.\\\int_{0}^{\pi} (\sin x+x\cos x)dx=\sin 2x\implies \sin 2x=0\implies 2x=\pi k\implies x=\dfrac{\pi}{2}k,\space k∈\mathbb{Z}.$$