CIN_U1_A6_LUHA
Transcript of CIN_U1_A6_LUHA
MATRICULA : AL12508085CARRERA: DESARROLLO DE SOFTWARE
45∫ 1
√X2−1=dx la integral de ∫ 1
√X2−1es cosh−1 ( x ) entonces=¿ ∫ 45
√X2−1=45cosh−1 ( x )+c
∫ √x dx=∫ x1/2dx=23 x3 /2+c
u=√x
y du= 1
2√ x
2∫(u2+1)du
2∫ (u2+1 )du=2∫u2du+2∫1du
2∫ (u2+1 )du=23u3+2u+c
23
[ x1/2 ]3+2√ x+c=23x3/2+2√ x+c
∫ senx
cosx2dx=∫ senx
(cosx ) (cosx )dx=∫ 1
cosxtanxdx=∫ secx tanx dx
Cambiando u=secx entonces du = tanx secx dx
∫1du=u+c regresandou asecx tenemos
∫ secx tanx dx=secx+c
∫ x3dx+∫ x1 /3dx+∫ 1
x3 /2dx=¿
34x4 /3+ x
4
4− 2
√x+c
∫ x2+x3+3x4
dx=∫( x2x4 + x3
x4+ 3x4 )dx∫( 1x2 + 1x + 3x4 )dx
∫( 1x2 + 1x + 3x4 )dx=∫ 1x2dx+∫ 1
xdx+3∫ 1
x4dx=¿
∫(x3+ 3√x+ 1x√ x )dx=−1
x3−1x+ log ( x )+c
∫ ( t3+3 t 4 )dt=∫ t 3dt+3∫ t 4dt
∫ t 3dt+3∫ t 4dt=3 t5
5+ t
4
4+c
∫10dz=10∫ dz=10 z+c
∫ (7 sen∅+cos∅ )d∅=7∫ sen∅ d∅+∫cos∅ d∅=7cos∅+sen∅+c
∫ sen∅1−¿¿
¿
∫ sen∅co s2∅
d