TALLER INTEGRAL 1. Resuelva cada una de las integrales usando el mΓ©todo de integraciΓ³n por partes: a. β« π₯ cos π₯ ππ₯. Sea
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TALLER INTEGRAL 1. Resuelva cada una de las integrales usando el mΓ©todo de integraciΓ³n por partes: a. β« π₯ cos π₯ ππ₯. Sea π’ = π₯ y ππ£ = cos π₯ ππ₯. Entonces ππ’ = ππ₯, π£ = sin π₯ y por tanto: β« π₯ cos π₯ ππ₯ = π₯ sin π₯ β β« sin π₯ ππ₯ = π₯ sin π₯ β (β cos π₯) + πΆ = π₯ sin π₯ + cos π₯ + πΆ. b. β« π₯ 2 ln π₯ ππ₯. 1 π₯
Sea π’ = ln π₯ y ππ£ = π₯ 2 ππ₯ entonces ππ’ = ππ₯ y π£ =
π₯3 . 3
Por tanto,
π₯3 π₯3 1 ln π₯ β β« ππ₯ 3 3 π₯ π₯3 1 = ln π₯ β β« π₯ 2 ππ₯ 3 3 π₯3 1 π₯3 = ln π₯ β +πΆ 3 3 3 π₯3 π₯3 = ln π₯ β + πΆ 3 9 3 π₯ = (3 ln π₯ β 1) + πΆ. 9
β« π₯ 2 ln π₯ ππ₯ =
c. β« π₯ 2 π π₯ ππ₯.
Sea π’ = π₯ 2 y ππ£ = π π₯ ππ₯. Entonces ππ’ = 2π₯ ππ₯ y π£ = π π₯ , por tanto, β« π₯ 2 π π₯ ππ₯ = π₯ 2 π π₯ β β« 2π₯π π₯ ππ₯ = π₯ 2 π π₯ β 2 β« π₯π π₯ ππ₯. Ahora tome π’ = π₯ y ππ£ = π π₯ ππ₯. Con esto se obtiene ππ’ = ππ₯, π£ = π π₯ y fialmente β« π₯ 2 π π₯ ππ₯ = π₯ 2 π π₯ β 2 (π₯π π₯ β β« π π₯ ππ₯) = π₯ 2 π π₯ β 2(π₯π π₯ β π π₯ ) + πΆ = π₯ 2 π π₯ β 2π₯π π₯ + 2π π₯ + πΆ. 2. Usando integraciΓ³n trigonomΓ©trica, resuelva cada integral: a. β« sin3 2π₯ cos 2 2π₯ ππ₯ β« sin3 2π₯ cos 2 2π₯ ππ₯ = β« sin2 2π₯ cos 2 2π₯ sin 2π₯ ππ₯
= β«(1 β cos2 2π₯) cos2 2π₯ sin 2π₯ ππ₯ Sea π’ = cos 2π₯ entonces ππ’ = β2 sin 2π₯ ππ₯ y, 1 β« sin3 2π₯ cos2 2π₯ ππ₯ = β β«(1 β π’2 )π’2 ππ’ 2
1 β« sin3 2π₯ cos2 2π₯ ππ₯ = β β«(π’2 β π’4 ) ππ’ 2 1 π’3 π’5 = β ( β )+πΆ 2 3 5 1 1 = β cos3 2π₯ + cos5 2π₯ + πΆ 6 10 sin π₯
b. β« (1βcos 2 ππ₯ π₯) Sea π’ = cos π₯, luego ππ’ = β sin π₯ ππ₯ y entonces, sin π₯ βππ’ ππ₯ = β« 2 (1 β cos π₯) (1 β π’)2
β«
Si π§ = 1 β π’ se sigue que ππ§ = βππ’ y se obtiene: β«
sin π₯ ππ§ ππ₯ = β« 2 = β« π§ β2 ππ§ = βπ§ β1 + πΆ. 2 (1 β cos π₯) π§
DevolviΓ©ndonos a la variable original, se tiene: β«
sin π₯ 1 ππ₯ = β +πΆ 2 (1 β cos π₯) 1βπ’ =
1 + πΆ. cos π₯ β 1
βtan π₯
c. β« sin π₯ cos π₯ ππ₯. βtan π₯ βtan π₯ ππ₯ = β« ππ₯ sin π₯ sin π₯ cos π₯ 2π₯ cos cos π₯ βtan π₯ =β« sec 2 π₯ ππ₯ tan π₯ 1 =β« sec 2 π₯ ππ₯ βtan π₯ Sea π’ = tan π₯ entonces ππ’ = sec 2 π₯ ππ₯. β«
β«
1 βtan π₯ ππ₯ = β« ππ’ sin π₯ cos π₯ βπ’ = 2β«
1
ππ’ 2βπ’ = 2βπ’ + πΆ = 2βtan π₯ + πΆ. 3. Aplicando el mΓ©todo de sustituciΓ³n trigonomΓ©trica, resuelva las siguientes integrales: 1
a. β« 2 2 2 ππ₯. π₯ βπ βπ₯ Sea π₯ = π sin π entonces ππ₯ = π cos π ππ y sustituyendo en la integral
β«
1 π₯ 2 βπ2 β π₯ 2
1
ππ₯ = β«
1
π cos π ππ π2 sin2 π βπ2 β π2 sin2 π 1 =β« π cos π ππ 2 2 π sin π πβ1 β sin2 π π cos π = 3β« 2 ππ π sin π πππ π 1 = 2 β« csc 2 π ππ π 1 = β 2 cot π + πΆ π 1 cos π =β 2 +πΆ π sin π 1 βπ2 β π₯ 2 =β 2 +πΆ π π₯
b. β« 2 2 ππ₯. π +π₯ Sea π₯ = π tan π, entonces ππ₯ = π sec 2 π ππ. AsΓ, 1 1 β« 2 ππ₯ = β« 2 π sec 2 π ππ 2 π +π₯ π + π2 tan2 π =
π sec 2 π β« ππ π2 1 + tan2 π
=
1 sec 2 π β« ππ π sec 2 π
=
1 β« ππ π
=
1 π + πΆ. π
π₯
Claramente π = arctan (π), por tanto, β« c. β«
π2
1 1 π₯ ππ₯ = arctan ( ) + πΆ. 2 +π₯ π π
βπ₯ 2 βπ2 ππ₯. π₯
Sea π₯ = π sec π, entonces ππ₯ = π sec π tan π ππ. Reemplazando en la integral, β«
βπ₯ 2 β π2 βπ2 sec 2 π β π2 ππ₯ = β« π sec π tan π ππ π₯ π sec π = β« πβsec 2 π β 1 tan π ππ = π β« tan π tan π ππ = π β« tan2 π ππ
Como tan2 π = sec 2 π β 1, se obtiene β«
βπ₯ 2 β π2 ππ₯ = π (β« sec 2 π ππ β β« ππ) = π(tan π β π) + πΆ π₯
Devolviendo a la variable original, β«
βπ₯ 2 β π2 βπ₯ 2 β π2 π₯ ππ₯ = π ( β arcsec ( )) + πΆ π₯ π π π₯ = βπ₯ 2 β π2 β π arcsec ( ) + πΆ π
4. Resuelva cada una de las siguientes integrales usando fracciones parciales π₯
a. β« π₯+1 ππ₯ β«
b. β«
π₯ π₯+1β1 ππ₯ = β« ππ₯ π₯+1 π₯+1 π₯+1 1 = β«( β ) ππ₯ π₯+1 π₯+1 1 = β« (1 β ) ππ₯ π₯+1 1 = β« ππ₯ β β« ππ₯ π₯+1 = π₯ β ln(π₯ + 1) + πΆ
π₯ 3 +1 ππ₯ π₯β2
β«
π₯3 + 1 π₯3 β 8 + 9 ππ₯ = β« ππ₯ π₯β2 π₯β2 (π₯ β 2)(π₯ 2 + 2π₯ + 4) β 9 =β« ππ₯ π₯β2 9 = β« (π₯ 2 + 2π₯ + 4 β ) ππ₯ π₯β2 π₯3 1 = + π₯ 2 + 4π₯ β 9 β« ππ₯ 3 π₯β2 π₯3 = + π₯ 2 + 4π₯ β 9 ln(π₯ β 2) + πΆ 3
π₯ 3 +1
c. β« (π₯+2)(π₯β1)3 ππ₯. π₯3 + 1 π΄ π΅ πΆ π· = + + + (π₯ + 2)(π₯ β 1)3 π₯ + 2 π₯ β 1 (π₯ β 1)2 (π₯ β 1)3 π₯ 3 + 1 = π΄(π₯ β 1)3 + π΅(π₯ + 2)(π₯ β 1)2 + πΆ(π₯ + 2)(π₯ β 1) + π·(π₯ + 2) π₯ 3 + 1 = (π΄ + π΅)π₯ 3 + (β3π΄ + πΆ)π₯ 2 + (3π΄ β 3π΅ + πΆ + π·)π₯ β π΄ + 2π΅ β 2πΆ + 2π· Por tanto, π΄+π΅ =1 β3π΄ + πΆ = 0 { 3π΄ β 3π΅ + πΆ + π· = 0 βπ΄ + 2π΅ β 2πΆ + 2π· = 1 Y resolviendo el sistema:
7 20 7 2 ,π΅ = ,πΆ = ,π· = . 27 27 9 3 Entonces la integral queda determinada por, π₯3 + 1 π΄ π΅ πΆ π· β« ππ₯ = β« ( + + + ) ππ₯ (π₯ + 2)(π₯ β 1)3 π₯ + 2 π₯ β 1 (π₯ β 1)2 (π₯ β 1)3 7 1 20 1 7 1 2 1 = β« ππ₯ + β« ππ₯ + β« + β« ππ₯ 2 (π₯ (π₯ 27 π₯ + 2 27 π₯ β 1 9 β 1) 3 β 1)3 7 20 7 1 = ln(π₯ + 2) + ln(π₯ β 1) β (π₯ β 1)β1 β (π₯ β 1)β2 + πΆ 27 27 9 3 7 20 7 1 1 1 = ln(π₯ + 2) + ln(π₯ β 1) β β +πΆ 27 27 9 π₯ β 1 3 (π₯ β 1)2 π΄=