Sustitución trigonometrica

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SUSTITUCION TRIGONOMETRICA

1. dx/√(9-x^2 )= ∫▒█(3cos/3cos dx= sin⁡θ^(-1) x/3+c=@@datos)
∫(3 cos⁡θ)/〖(3 sin⁡〖θ)〗〗^2 3 cos⁡〖θ 〗 dθ=∫(9 〖(cos〗⁡〖θ)〗^2)/(9 〖(sin〗⁡〖θ)〗^2 ) ∫( 〖(cos〗⁡〖θ)〗^2)/( 〖(sin〗⁡〖θ)〗^2 ) ∫〖(cot〗⁡〖θ)〗^2 ∫〖(csc〗⁡〖θ)〗^2 dθ-∫1dθ
∫▒dθ= θ+C
datos
√(9-x^2 )=3 cos⁡θ
x=3 sin⁡θ
dx=3 cos⁡θ

2.∫▒〖(∫▒√(25-x^2 ))/x dx=∫▒〖(5 cos⁡5cos)/5sen=5∫▒〖〖cos〗^2/sen dx=5∫▒〖1-sen〗^2/sen=5∫▒〖1/sin⁡θ =-〖sen〗^2/sen2 dx〗〗〗〗
=5[∫▒〖csc-∫▒sen〗]=5[█(-ln⁡〖|csc+cot⁡|+cos〗@)]
=-ln⁡〖|5/x+ √(25-x^2 )/x|+√(25-x^2 )/5〗
=-5 ln⁡〖|5/x+√(25-x^2 )/x|+5 √(25-x^2 )/5+c〗
Datos
a=25 aa^2=5
a=cos
x=5 sen
dx=5cos

3. ∫▒〖√(x^2-4dx) 〗= 4∫▒〖〖tan〗^2 sec⁡〖=4∫▒〖〖sec〗^3-sec⁡〖=4∫▒〖〖sec〗^3 ∫▒sec⁡〖=4∫▒〖〖sec〗^2 sec〗-sec〗〗 〗 〗〗 〗
4∫▒〖〖(tan〗^2+1)sec-∫▒sec〗 =4( 1/2 secθtanθ+1/2 ln⁡|sec+tanθ|-ln⁡|secθ+tanθ+c
Datos
a^2=4
a=2
x= 2sec
x/5=sen
dx=2sec tan

4.∫▒dx/(x^2-1)^3 =∫▒〖=sec⁡tan/〖tan〗^3 〗= ∫▒sec/〖tan〗^2 =∫▒cos/〖sen〗^2 u=sen du= cos
∫▒〖du/u^2 =∫▒u^(-2) du u^(-1)/(-1)+c=-(sen)^(-1)+c=1/sen+c〗
x/1=sec -1/(√(x^2-1)/x)+c=(-x)/√(x^2-1)+c
Datos
X=sec θ dx=secθtanθ dθ



5.∫▒((1-x^2 )^(3⁄2))/x^6 dx ∫▒〖=√((1-x)^3 )/x^6 〗 dx
∫▒〖(〖cos〗^3 θcosθ)/(〖sen〗^6 θ) dθ〗=∫▒cos4θ/(〖sen〗^6 θ) u=senθ du=cosθ
∫▒〖cos〗^4/〖sen〗^4 =1/〖sen〗^2 dθ∫▒〖〖cotθ〗^4 csc^2 θdθ=-∫▒〖u^4 du〗〗=u^4/5+c=-(cot^5 θ)/5+c
=(√(〖1-x〗^2 )/x)/(5/1)+c=-((√(1-x^2 )^5 ))/5x+c
Datos
X=a senθ
Dx=cosθ

FRACCIONES PARCIALES
1.∫▒(3X-2)/(X3-x^2-2X)=∫▒〖(3X-2)/X(X-2)(X+1) =〗 ∫▒〖A/X+B/(X-2)+C/(X+1)〗
X^3-x^2-2X=X(x^2-X-2)=X(X-2)(X+1)
=∫▒(A(X-2)(X+1)+B(X)(X+1)+C(X)(X-3))/X(X-2)(X+1)
=(3X-2)/(X(X-2)(X+1))=(A(X-2)(X+1)+B(X)(X+1)+C(X)(X-2))/(X(X-2)(X+1))
A(x^2+X-2-2)+B(x^2+2)+C(x^2-2X)
3X-2=AX^2-AX-2A+Bx^2+BX+C^2-2CX
3X-2=x^2 (A+B+C)+X(-A+B-2C)-2A
x^2=A+B+C=0
X=-A+B-2C=3
CONSTANTE=-2A=-2
-2A=-2
A=2/(-2)=1
EVALUAREN 1 Y2
1+B+C=0 B+C=1 2B+2C=-2
C-1+B-2C=3 B-2C=4 B-2C=4
3B=2 C=-5/3
=∫▒〖1/X DX+∫▒〖(2/3)/(X-2) DX+∫▒〖((-5)/3)/(X+1) DX 〗〗〗=ln⁡|x|+2/3 ln⁡|x-2- 5/3 ln⁡|x+1|
=ln|x(x-2)^(2⁄3)-ln⁡(x+1)^(5⁄3)=ln⁡|(x(x-2〖_)〗^(5⁄3))/((x+1)^(5⁄3) )|+c

2.∫ dx/(5x+x^2 ) ∫A/x+B/(x+5) (A(x+5)+b(x))/x(x+5) =dx/(5x+x^2 ) A(x+5)+Bx
Factorizar
5x+x^2=x(x+5)
x=0 x+5=0x=-5

x=0
A(x+5)+Bx=1
A(0+5)+B0=1
A(5)=1
A=1/5
x=-5
A(-5+5)+B(-5)=1
A(0)+B(-5)=1
B(-5)=1
B=-1/5
∫(1⁄5)/x+∫((-1)⁄5)/(x+5)
1/5∫1/x-1/5∫1/(x+5)
1/5 ln⁡x-1/5 ln⁡〖x+5+C〗

3.∫▒〖(2x-1)/(x(x^2+3x+2)) dx=∫▒〖a/x+b/((x+2) )+c/((x+1) ) 〗〗
Factorizar
X(x+2)(x+1) ∫▒(a(x+2)(x+1)+b(x)(x+1)+c(x)(x+1))/(x(x+2)(x+1))(2x-1)/(x(x+2)(x+1))=(a(x^2+3x+2)b(x^2+x)+c(x^2+2x))/(x(x+2)(x+1))
2x-1=ax^2+3ax+2a+bx^2+bx+cx^2+2cx 2x-1=x^2 (a+b+c)+x(3a+b+2c)2a
x^2=a+b+c=0 x=3a+b+2c=2 cte=2a=-1
A=(-1)/2-1/2+b+c=0 b+c=1⁄2 b+3=1/2 b=1/2-3 b=(-5)/2
(█(-1/2+b+c=0@3/2-b-2c=-2) )/█(1-c=-2@xc=-3) ∫▒〖(-1⁄2)/x+∫▒〖(-5⁄2)/(x-2) dx+∫▒〖3/(x+1) dx= -1/2 ∫▒〖1/x dx-5/2 ∫▒█(1/((x-2) ) dx+3∫▒〖1/(x+1)dx〗@)〗〗〗〗
-1/2 ln⁡|x|-5/2 ln⁡|x-2|+3|n|x+1|+c =ln⁡|((x+1)^3)/((x+2)^(5⁄2) (x)^(1/2) )|+c

4.∫▒(x^4-x^3-x-1)/(x^3-x^2 ) dx =∫▒- a/x+b/x^2 +c/(x-1)
Factorizar
x^3-x^2=x^2 (x-1)
.∫▒(x^4-x^3-x-1)/(x^3-x^2 )=(a(x)(x-1)+b(x-1)+c(x^2))/(x^2 (x-1))
x+1=ax^2-ax+bx-b+cx^2
x+1=x^2 (a+c)+x(-a+b)-b
x^2:a+c=0
X:-a+b=1
Ctes-b==1
A=-2 b=-1 c=2
x^2/2+lnx^2/((x-1)^2 )-1/x+c

5.∫▒〖(2x^2-25x-23)/((x+1)^2 (x-5)) ∫▒█(a/((x+1) )+b/((x+1)^2 )+c/(x-5dx ) x+1=0 x+1^2=0 x-5=0@)〗
(a(x+1)(x-5)+b(x-5)+c(x+1)^2)/((x+1)^2 (x-5))
=ax^2-4ax-5a+bx-5b+cx^2+2cx+c
x^2:a+c=2
X:-4a+b+2c=-25
Cte:-5ª-5b+c=-33
a=5 c=-3 b=1
∫▒〖5/((x+1) )+∫▒〖1/(x+1)^2 dx+∫▒〖(-3)/(x-5) dx〗〗〗
Ln|〖(x+1)〗^5/〖x+5〗^3...
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