Integrales resueltas schaum

Solo disponible en BuenasTareas
  • Páginas : 15 (3657 palabras )
  • Descarga(s) : 0
  • Publicado : 26 de noviembre de 2010
Leer documento completo
Vista previa del texto
74.- ∫▒e^tg2x 〖sec〗^(2 ) 2x dx= 1/(2 ) ∫▒e^(tg2x ) 〖sec〗^(2 ) 2x dx= 1/(2 ) e^(tg2 )+ c
v=tg 2x
dv= 〖sec〗^2 2x .2dx

75.-∫▒e^(2sen 3x ) cos⁡〖3x dx= 1/6〗 ∫▒e^(2sen 3x ) 2 cos⁡〖3x .3dx= 1/6〗 e^(2sen 3x )+ c
e^(v )= e^(2sen 3x)
v=2sen 3x
dv=2cos⁡〖 3x.dx 〗

76.-∫(dx )/√(5-x^2 ) = ∫▒(dv )/√(a^(2 )–v^2 ) =arc sen v/a +c=arc sen x/√5+ c=
arc sen (x√5)/5+ c
a^(2 )=5 ;a=√5
v^2= x^2
v=dx
dv=dx

77.- ∫▒dx/(5+x^(2 ) ) = ∫▒dv/(a^2+v^2 ) = ∫▒dv/(v^2+a^2 ) = 1/a arc tg v/a + C = 1/√5 arc tg x/(√5 ) + C = √5/5 arc tg (x√5 )/5 + C
v^2=x^2;v=x ;dv=dx ; a^2=5;a=√5

78.- ∫▒(dx )/(x√(x^2 )- 5 )=∫▒dv/(v√(v^2-a^2 )) = 1/(a ) arc sen v/(a )+ c
1/(√5 ) arc sen x/√5+ C = √5/5 arc sen (x√5)/5+ C
v^(2 )= x^(2 );v=x ;dv=dx ; a^2=5;a= √579.- ∫▒(e^x dx)/√(1-e^2x )= ∫▒dv/√(a^2-v^2 )=arc sen v/a+ C=arc sen e^x+ C
v^(2 )= e^2x ;v=e^x ;dv= e^x dx; a^2=1 ;a=1

80.-∫▒〖(e^2x dx)/(1+〖e 〗^4x )= 1/2 ∫▒(e^2x 2dx)/(1+ e^4x )〗=1/2 ∫▒〖dv/(a^2+ v^2 )=1/2 ∫▒〖dv/(v^2+a^2 )=1/2 1/a arc tg v/a+ C 〗〗
1/2 arc tg e^2x+ C
v^2= e^4x;v=e^2x ;dv=e^2x.2dx ;a^2=1 ;a=1

81.- ∫▒dx/√(4-9x^2 )= 1/3 ∫▒3dx/√(4-9x^2 )= 1/3 ∫▒dv/√(a^2-v^2 ) = 1/3arc sen v/a+ C= (1 )/3 arc sen 3x/2+C
v^2=9x^2;v=3x;dv=3dx;a^2=4;a=2
82.- ∫▒〖dx/(9x^(2 )+4)= 1/3〗 ∫▒3dx/(〖9x〗^2+4)= 1/3 ∫▒〖dv/(v^2+a^2 )= 1/3 1/a arc tg v/a〗 + C = 1/6 arc tg 3x/2+C
v^2=〖9x〗^2;v=3x;dv=3dx;a^2=4;a=2

83.- ∫▒(sen 8x)/(9+〖sen〗^4 4x) dx= 1/4 ∫▒(sen 8x 4dx)/(9+〖sen〗^4 4x)=1/4 ∫▒dv/(a^2+v^2 )= 1/4 1/a arc tg v/a+C
1/12 arc tg (〖sen〗^2 4x)/3+C
v^2= 〖sen〗^4 4x ;v=〖sen〗^24x ;dv=2sen 4x cos4x 4dx
dv=sen 8x 4dx ; a^2=9;a=3

84.- ∫▒(〖sec〗^2 xdx)/√(1-4〖tg〗^2 x)= 1/2 ∫▒(2〖sec〗^2 xdx)/√(1-4〖tg〗^2 x)=1/2 ∫▒dv/√(a^2-v^2 ) = 1/2 arc sen v/a+C=
1/2 arc sen (2tgx)+C
v^2=4〖tg〗^2 x; v=2tgx;dv=2〖sec〗^2 xdx; a^2=1;a=1
85.-∫▒〖dx/(x√4-9〖ln〗^2 x)=∫▒〖1/(√4-9〖ln〗^2 x).dx/x=1/3 ∫▒〖3/√(4-9〖ln〗^2 ) dx/x=〗〗〗
1/3 ∫▒〖dv/√(a^2-v^2 )=1/3 arc sen v/a+ C=1/3 arc sen 3/2 ln⁡〖x+C=〗〗
=1/3 arc sen ln x^(3/2)+ C
v^2=9〖ln〗^2 x ;v=3ln⁡〖x;dv=3 dx/x;a^2=4;a=2〗
86.-∫▒〖(〖2x〗^4-x^2)/(〖2x〗^2+1) dx =∫▒〖x^2 dx-∫▒〖dx+∫▒〖dx/(〖2x〗^2+1)=∫▒〖v^n dv-∫▒〖dx+1/√2 ∫▒du/(u^2 a^2 )〗〗〗〗〗〗=
=v^(n+1)/(n+1)-x+1/√2 1/a arc tg u/a+ C=x^3/3-x+1/√2 arc tg x√(2 )+ C=
=1/3 x^3-x+√2/2 arc tg x√(2 )+ C
v=x ;n=2 ;dv=dx;
u^2=〖2x〗^2;u=x√2 ;du=√2 dx;a^2=1;a=1
87.-∫▒〖cos⁡〖2x dx〗/(〖sen〗^2 2x+8)=1/2∫▒〖cos⁡〖2x 2dx〗/(〖sen〗^2 2x+8)=1/2 ∫▒〖dv/(v^2+a^2 )=1/2 1/a arc tg v/a+ C=〗〗〗
=1/(2√8) arc tg (sen 2x)/√8+ C
=√8/16 arc tg (sen 2x)/(2√8)+ C=√((4)(2) )/16 arc tg (sen 2x)/√((4)(2) )+ C=
=√2/8 arc tg (sen 2x)/(2√2)+ C
88.-∫▒〖(2x-3)dx/(x^2+6x+13)=∫▒(2x-3+6-6)dx/(x^(2 )+ 6x+13)==∫▒〖((2x+6)dx-9)/(x^2+6x+13)=∫▒〖(2x+6)dx/(x^2+6x+13)-9∫▒〖dx/(x^2+ 6x+13)=〗〗〗〗
=∫▒dv/v-9∫▒〖dx/(〖(x+3)〗^2+4)=∫▒〖dv/v-9∫▒du/(u^2+a^2 )〗〗=ln⁡v-91/a arc tg u/a+C=

=ln⁡(x^2+ 6x+13)-9/2 arc tg|(x+3)/2|+ C
v=x^2+6x+13;dv=(2x+6)dx
u^2=〖(x+3)〗^2;u=x+3;du=dx;a^2=4;a=2
89.- ∫▒(x-1)dx/(〖3x〗^2-4x+3)= 1/6 ∫▒(6x-6)dx/(〖3x〗^2-4x+3) = 1/6 ∫▒█((6x-6-4+4)dx/(〖3x〗^2-4x+3)@ )
1/6 ∫▒〖(6x-4)dx/(〖3x〗^2-4x+3)-1/3 ∫▒〖dx/(〖3x〗^2-4x+3)=1/6 ∫▒〖dv/v-1/3 ∫▒█(dx/(〖3x〗^2-4x+3)=@ )〗〗〗
1/6 ∫▒〖dv/v-1/3 ∫▒dx/(〖[x √3-2/√3]〗^2+5/3)〗=1/6ln⁡(〖3x〗^2-4x+3)-1/(3√3) 1/(√5/√3) arc tg (x√3-2/√3)/(√5/√3)+C=
=1/6 ln⁡(〖3x〗^2-4x+3)-√3/(3√3 √5) arc tg ((3x-2)/√3)/(√5/√3)+C=
=1/6 ln(〖3x〗^2-4x+3)-√5/15 arc tg (3x-2)/√5+C
Variables: v=〖3x〗^2-4x+3;dv=(6x-4)dx; u^2=(x√3-2/√3)^2;u(x√3-2/√3);du√3 dx;
a^2=5/3 ;a=√(█(5/3@ ))
90.-∫▒〖xdx/(√27+6x-x^2 )=∫▒█((27+6x-x^2 )^(-1/2) xdx=@ )〗
-1/2 ∫▒〖〖(27+6x-x^2)〗^(-1/2) (-2x+6) 〗 dx+3∫▒█(dx/√(27+6x-x^2 )=@ )-1/2 ∫▒〖v^n dv+3〗 ∫▒〖dx/√(-(x^2-6x-27) )=-1/2 ∫▒〖v^n dv+3∫▒〖dx/(√36-〖(x-3)〗^2 )=〗〗〗
-1/2 ∫▒〖v^n dv+3∫▒〖du/√(a^2-v^2 )=-√(27+6x-x^2 ) 〗〗+ 3 arc sen ⌈(x-3)/6⌉+c

Variables:v=(27+6x-x^2 ); dv=(6-2x)dx ;n= -1/2
u^2=〖(x-3)〗^2;u=(x-3);du=dx; a^2=36;a=6

91.-∫▒〖(5-4x)dx/√(12x-〖4x〗^2-8)=∫▒〖(12x-〖4x〗^2-8)^(-1/2) (5-4x)dx=〗〗
1/2 ∫▒〖(12x-〖4x〗^2 〗-〖8)〗^(-1/2) (10-8x)dx=
1/2...
tracking img