INTEGRAL EULERIANA SOLUCIONES 4
π/2 (2p-1=3 , 2q-1=0) Γ(2) · Γ(1/2) Γ(1/2) 2
1) sen3x dx = 1/2 · β(2,1/2) = = =
2 · Γ(5/2) 2 · 3/2 · 1/2 · Γ(1/2) 3
0
π/2 (2p-1=3 , 2q-1=2/3) Γ(2) · Γ(5/6) Γ(5/6) 36
2) 2 sen3x cos2/3x dx = β(2,5/6) = = =
Γ(17/6) 11/6 · 5/6 · Γ(5/6) 55
0
π/2 π/2 (2p-1=-1/2 , 2q-1=1/2) Γ(1/4) · Γ(3/4)
3) √(1/tg x) dx = sen-1/2x cos1/2x dx = 1/2·β(1/4,3/4) = = 1/2·Γ(1/4)·Γ(3/4)
2 · Γ(1)
0 0
1 Cambio Var. 1 (p-1=3 , q-1=-1/2) 4·Γ(4)·Γ(1/2) 128
4) (1-x1/4)-1/2 dx (x1/4= t) 4 t3 (1-t)-1/2 dt = 4·β(4,1/2) = =
Γ(9/2) 35
0 0
1-√x Cambio Var. (p-1=3/2) 3·e·√π
5) x1/4 e (√x = t) 2·e e-t t3/2 dt = 2·e·Γ(5/2) = 2·e·3/2·1/2·Γ(1/2) =
2
0 0
(p-1=-2/3)
6) e-x / x2/3 dx = Γ(1/3)
0
-x/4 Cambio Var. (p-1=3)
7) x3 e dx (x/4 = t) 44 e-t t3 dt = 44 · Γ(4) = 1.536
0 0
1 L x Cambio Var. (p-1=1)
8) dx (x = e-2 t) 2 e-t t  dt = 2 · Γ(2) = 4
√x
0 0
1 Cambio Var. (p-1=4)
9) L4x dx (x = e- t) e-t t 4 dt = Γ(5) = 4! = 24
0 0
1 x 1 (p-1=1/2 , q-1=-1/2) Γ(3/2) · Γ(1/2) π
10) √( ) dx = x1/2 (1-x)-1/2 dx = β(3/2,1/2) = = 1/2 · Γ2(1/2) =
1-x Γ(2) 2
0 0
1 dx Cambio Var. 1 1 (p-1=-1/2 , q-1=-1/3) Γ(1/2)·Γ(2/3) 3√π·Γ(2/3)
11)     (x2 = t) t-1/2(1-t)-1/3dt = 1/2·β(1/2,2/3) = =
3√(1-x2) 2 2 · Γ(7/6) Γ(1/6)
0 0
3 dx Cambio Var. 1 (p-1=-1/2 , q-1=-1/2)
12) (3-x = t) t-1/2(1-t)-1/2dt = β(1/2,1/2) = π
√(3-x)(x-2)
2 0
1 dx Cambio Var. 1 1 (p-1=-3/4 , q-1=-1/2) Γ(1/4)·Γ(1/2) √π·Γ(1/4)
13)     (x2 = t) t-3/4(1-t)-1/2dt = 1/2·β(1/4,1/2) = =
√ x (1-x2)
2 2 · Γ(3/4) 2·Γ(3/4)
0 0
π/2 cos x  dx Cambio Var.
14) (sen x = t)
√(1-sen x)
0