Matematika/Furje eilutės: Skirtumas tarp puslapio versijų

:čia <math>\text{d}((k+n)x)=(k+n)\text{d}x, \; \frac{\text{d}((k+n)x)}{k+n}=\text{d}x</math> ir <math>\text{d}((k-n)x)=(k-n)\text{d}x, \; \frac{\text{d}((k-n)x)}{k-n}=\text{d}x</math> bei pasinaudojome trigonometrine formule <math>\sin(A)\cdot\sin(B) = -\frac{1}{2}[\cos(A + B) - \cos (A - B)];</math>

:<math>\int_{-\pi}^{\pi}\sin(kx)\cos(nx)\text{d}x=\frac{1}{2}\int_{-\pi}^{\pi}[\sin((k+n)x)+\sin((k-n)x)]\text{d}x=-\frac{1}{2}\Big(\frac{\cos((k+n)x)}{k+n}+\frac{\cos((k-n)x)}{k-n}\Big)|_{-\pi}^\pi=0. \quad (5.2)</math>

:<math>\int_{-\pi}^{\pi}\sin(px)\cos(px)\text{d}x=\int_{-\pi}^{\pi}\sin(px)\cos(px)\frac{\text{d}(\sin(px))}{p\cos(px)}=\frac{1}{p}\int_{-\pi}^{\pi}\sin(px)\text{d}(\sin(px))=\frac{1}{2p}\sin^2(px)|_{-\pi}^{\pi}=0</math> čia <math>\text{d}(\sin(px))=p\cos(px) \; dx.</math>

:Pagaliau,