La integración por partes
Enviado por shane11 • 26 de Mayo de 2013 • Tareas • 291 Palabras (2 Páginas) • 443 Visitas
INTEGRACION POR PARTES
=xsenx+cosx+C
=xarc cos〖2x- 1/2 √(1-4x^2 )〗+C
= -2/105 (1-x)^(3/2) (15x^2+12x+8)+C
= 1/2 (x^2+1) arctanx- 1/2 x+C
= -2/3 cos^3 x-sen^2 x cosx+C
= (2(bx-2a)√(a+bx))/(3b^2 )+C
= 1/2 x^2 arc sen x^2+ 1/2 √(1-x^4 )+C
= 1/2 x(sen lnx-coslnx)+C
= (e^ax (a senbx-b cosbx))/(a^2+b^2 )+C
=x/(4(4+x^2 )^(1/2) )+C
= 1/2048 {(x(3x^2-80))/((x^2-16)^2 )+3/8 ln|(x-4)/(x+4)|}+C
= 3/8 x-3/8 senx cosx-1/4 sen^3 xcosx+C
= -1/5 cos^3 x(sen^2 x+2/3)+C
INTEGRALES TRIGONOMETRICAS
∫▒〖cos^2 x dx= 1/2 x+ 1/4 sen 2x+C〗
= 1/6 cos^2 2x- 1/2 cos2x+C
∫▒〖〖(sen 2x)〗^4 dx=3/8 x-1/8 sen 4x〗+1/64 sen8x+C
= 3/8 x+1/2 senx+ 1/16 sen2x+C
∫▒〖〖(sen x)〗^7 dx=1/7 〖(cosx)〗^7-8/5 (cosx)^5+(cosx)^3-cosx+C〗
=5/16 x+1/2 senx+3/32 sen2x-1/24 sen^3 x+C
∫▒〖(senx)^2 (〖cosx)〗^5 dx=1/3 (senx)^3-2/5(〖senx)〗^5+1/7 (senx)^7+C〗
= 1/5 cos^5 x-1/3 cos^3 x+C
∫▒〖〖(senx)〗^3 〖(cosx)〗^3 dx=1/48 〖(cos2x)〗^3-1/16 cos2x+C〗
=1/128 (3x-sen4x+ 1/8 sen8x)+C
∫▒〖sen2x cos4x dx=1/4 cos2x-1/12 cos6x+C〗
= 1/2 senx+ 1/10 sen5x+C
∫▒〖sen 5x sen x dx=1/8 sen 4x-1/12 sen 6x+C〗
=senx+ 1/2 sen^2 x+C
∫▒〖〖(cosx)〗^(2⁄3)/〖(senx)〗^(8⁄3) dx〗 = -3/5 〖cot〗^(5⁄3) x+C
=csc〖x- 1/3 csc^3 x+C〗
∫▒〖x(cos^3
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