prÁctica calificada.docx
TRANSCRIPT
PRÁCTICA CALIFICADA
1. ∫( x2 lnxx
¿)dx ¿
∫ x2
x+∫ ln xx
∫ xdx+¿∫ ln xx ¿
12x+ ln
2 x2
+c
12x+ ln x+c
2. ∫ x23√1+2x
dx
1+2x = u3
x = u3−12
∫ x23√1+2x
dx = ∫( u
3−12
)2
3√u3. 3u
2du2
= 32∫u6−2u3+1
4u.u2du
2dx = 3u2du
dx = 3u2du2
38∫ (u7−2u4+u )du = 38 u
8-34. u5
5+38 . u
2
2 + C (debemos a la variable x)
= 38(1+2x )
83 - 320
(1+2x )53+ 316
(1+2x )23 + C
3. tan3 x sec xdx
u = secx du = secxtanx dx (sec(x¿¿2- 1) = tan(x2)
integral de tan3(x).sec(x)dx = integral de tan(x).(sec(x2- 1).sec(x)dx ay que integrar u2-1 du u33- u + c reemplzar donde estaba u sec(x)
y luego:
sec3x3 -secx +c
4. ∫ sen5 x cos2 xdx∫ sen5 xcos2xdx = ∫ sen4 x cos2 xsenxdx
= ∫(sen¿¿2 x )2cos2 xsenxdx ¿
= ∫¿¿xsenxdx
= ∫(1−2cos2 x+cos4 x )cos2 xsenxdx
= ∫cos2 xsenxdx−2∫ cos4 xsenxdx+∫cos6 xsenxdx….(I) Si hacemos: cosx = u −senxdx=du senxdx=−du
Reemplazamos en (I): = ∫u (−du )−2∫ u4 (−du )+∫ u6(−du) = −∫udu+2∫ u4du−∫ u6du = −12 u
2+ 25u5−1
7u7+C →Volvemosa la variable
∫ sen5 xcos2 xdx = −12 cos2 x+2
5cos5 x−1
7cos7x+C Rpta.
5. ∫cos5 xdx ∫cos4 xcosxdx =∫(cos¿¿2x )2 cosxdx=∫(1−sen2x)2 cosxdx ¿
= ∫(1−2 sen2¿x+sen4 x )cosxdx¿
= ∫ cosxdx−2∫ sen2 xcosxdx+∫ sen4 xcosxdx ∫cos5 xdx = senx−2
3sen3 x+ 1
5sen5 x+C Rpta.
6. ∫ sec5 xdx
∫ sec^5 x dx = ∫ sec3 sec2 x dx = tanx sec2 x - ∫ tan x (3 sec2 x sec x tan x) dx = tanx sec2 x - 3 ∫ tan^2 x sec3 x dx = tanx sec2 x - 3 ∫ (sec2 x -1) sec3 x dx = tanx sec2 x - 3 ∫ sec5 x dx + 3 ∫ sec3 x dx
Resulta que
4 ∫ sec5x dx = tanx sec2 x + 3 ∫ sec3 x dx
∫ sec2 x dx = ∫ sec x sec2 x dx = tan x sec x - ∫ tan x sec x tanx dx = tan x sec x - ∫ (sec2 x -1) sec x dx = tan x sec x - ∫ sec3 x dx + ∫ sec x dx
Resulta que
2 ∫ sec3 x dx = tan x sec x + ln(sec x + tan x)
∫ sec3x dx = 1/2(tan x sec x + ln(sec x + tan x)).
Finalmente
4 ∫ sec5 x dx = tanx sec2x + 3 ∫ sec^3 x dx
∫ sec5 x dx = 1/4(tanx 2 x + 3 ∫ sec3x dx)
14 (tan x sec2 x + 3/2 (tan x sec x + ln(sec x + tan x)) + C. Rpta.
7. ∫3sin h (7 x )−8cos h (7 x )dx
∫ (3sinh (7 x )−8cosh(7x )dx )
3∫ sinh (7 x )dx−8∫ cosh (7 x)dx
3cosh (7 x)7−8 sinh(7 x)
7+C
8. ∫(x5¿+ x4+x3+x2−1)℮2 xdx ¿
∫(x5¿+ x4+x3+x2−1)℮2 x¿=℮2x
2 [ x5+ x4+x3+x2−1−5 x4+4 x3+3x2+2 x2+20x3+12x2+6 x+2
4−60x2+24 x+6
8+120 x+2416
+12032 ]
+C
∫(x5¿+ x4+x3+x2−1)℮2 x¿=℮2x
2 [ x5+ x4+x3+x2−1−5 x4+4 x3+3x2+2 x2+10 x3+6 x2+3 x+1
2−30x2+12x+3
4+15x+32
+154 ]
+C