Ejercicios De Integrales De Funciones Trigonometricas(1).pdf
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Ejercicios importantes Ejercicio 1. β« ππ¨π¬ π π π
π 2
β« cos 4 π₯ ππ₯ = β« (cos 2 π₯) ππ₯ 1 + cos 2 π₯ 2 = β«( ) ππ₯ 2 1 = β« 1 + 2cos 2π₯ + cos 2 2π₯ ππ₯ 4 1 1 1 = β« ππ₯ + β« 2cos 2π₯ππ₯ + β« cos 2 2π₯ ππ₯ 4 4 4 1 1 1 1 + cos 4π₯ = β« ππ₯ + β« 2cos 2π₯ππ₯ + β« 4 4 4 2 1 1 1 1 = β« ππ₯ + β« 2cos 2π₯ ππ₯ + β« ππ₯ + β« cos 4π₯ ππ₯ 4 4 8 8 1 2π ππ 2π₯ 1 π ππ 4π₯ = (π₯ + ) + (π₯ + )+πΆ 4 2 8 4 1 1 1 1 = π₯ β π ππ 2π₯ + π₯ β π ππ 4π₯ + πΆ 4 4 8 32 3 1 1 = π₯ β π ππ 2π₯ + π ππ 4π₯ + πΆ 8 4 32 Ejercicio 2. β«
β«
π π πππππ
π π = β« π ππ 6 π₯ππ₯ ; ππππ π = β« π ππ 4π₯π ππ 2π₯ππ₯ ; = β«(π ππ 2π₯)2 π ππ 2π₯ππ₯ ; = β«(1 + π‘ππ2 π₯)2 π ππ 2π₯ππ₯ ;
π‘ππ2 π₯ + 1 = π ππ 2 π₯
= β«(π‘ππ4 π₯ + 2π‘ππ2 π₯ + 1) π ππ 2π₯ππ₯ ; = β«(π‘ππ4 π₯ π ππ 2 π₯ + 2π‘ππ2 π₯ π ππ 2π₯ + π ππ 2 π₯ ) ππ₯ ; = β« π‘ππ4 π₯ π ππ 2 π₯ ππ₯ + β« 2π‘ππ2 π₯ π ππ 2 π₯ ππ₯ + β« π ππ 2 π₯ ππ₯ ; = β« π‘ππ4 π₯ π(π‘πππ₯) + 2 β« π‘ππ2 π₯ π(π‘πππ₯) + π‘πππ₯ ; =
1 1 π‘ππ5 π₯ + π‘ππ3 π₯ + tan π₯ + πΆ ; 5 3
Ejercicio 3. β« ππππ ππ π π
β« ππππ ππ π π = β« πππ‘ 2 3π₯(πππ‘ 2 3π₯) ππ₯ ; = β« πππ‘ 2 3π₯(ππ π 2 3π₯ β 1)ππ₯ ; π’ = cot 3π₯ ππ’ = β csc 2 3π₯(3)ππ₯ ππ’ ππ₯ = β 3csc 2 3π₯
= β« cot 2 3 π₯ ππ π 2 3π₯ππ₯ β cot 2 3 π₯ ππ₯ ; = β« cot 2 3 π₯ ππ π 23π₯ππ₯ β β« cot 2 3π₯ ππ₯ ; = β« π’2 csc2 3π₯ (β
ππ’ 3csc2 3π₯
) β β« csc2 3π₯ β 1 ππ₯ ;
1 = β β« π’2 ππ’ β β« csc2 3π₯ + β« ππ₯ 3 =β
1 π’3 cot 3π₯ )+π₯ β (β 3 3 3
=β
π’3 1 + cot 3π₯ + π₯ + πΆ ; 9 3
=β
πππ‘ 3 3π₯ 1 + cot 3π₯ + π₯ + πΆ ; 9 3
Ejercicio 4. β« ππππ π π π
β« ππππ π π π = β«(πππ‘ 2 π₯ + 1)2 (ππ π 2 π₯) ππ₯ ;
π’ = cot π₯ ππ’ = β csc 2 π₯ ππ₯ ππ’ ππ₯ = β csc 2 π₯
= β«(πππ‘ 4 π₯ + 2 cot 2 π₯ + 1)(ππ π 2 π₯)ππ₯ ; = β« cot 4 π₯ ππ π 2 π₯ + 2 cot 2 π₯ππ π 2 π₯ + csc2 π₯ ππ₯ ;
= β« cot 4 π₯ ππ π 2 π₯ππ₯ + β« 2 cot 2 π₯ππ π 2 π₯ππ₯ + β« csc2 π₯ ππ₯ ; = β« π’4 ππ π 2π₯ (β
ππ’ csc2
) + 2 β« u2 π₯ππ π 2 π₯ (β
π₯
= β β« π’4 ππ’ β 2 β« u2 ππ’ + β« csc2 π₯ ππ₯ ; π’5 2π’3 =β β β cot π₯ + πΆ 5 3 πππ‘ 5 π₯ 2πππ‘ 3 π₯ =β β β cot π₯ + πΆ 5 3
ππ’ csc2
) + β« csc2 π₯ ππ₯ ;
π₯
Ejercicio 6. β« ππππ πππππ π π π
β« ππππ πππππ π π π = β« tan4 π₯ π ππ 6 π₯ sec π₯ tan π₯ ππ₯ ; = β«(sec2 π₯ β 1)2 sec6 π₯ π πππ₯ tan π₯ ππ₯ ; = β«(sec4 π₯ β 2 sec2 π₯ + 1) sec6 π₯ π πππ₯ tan π₯ ππ₯ ; = β« (sec10 π₯ β 2 sec8 π₯ + sec6 π₯) π πππ₯ tan π₯ ππ₯ ; = β« (u10 β2 u8 + u6 ) π πππ₯ tan π₯ π’11 2π’9 π’7 = β + +πΆ 11 9 7 π ππ 11π₯ 2π ππ 9 π₯ π ππ 7π₯ = β + +πΆ 11 9 7
ππ’ sec π₯ π‘πππ₯
;
π’ = sec π₯ ππ’ = π πππ₯ tan π₯ ππ₯ ππ’ ππ₯ = sec π₯ π‘πππ₯
β« cos 4 π₯ ππ₯ = β« (cos 2 π₯) ππ₯ 1 + cos 2 π₯ 2 = β«( ) ππ₯ 2 1 = β« 1 + 2cos 2π₯ + cos 2 2π₯ ππ₯ 4 1 1 1 = β« ππ₯ + β« 2cos 2π₯ππ₯ + β« cos 2 2π₯ ππ₯ 4 4 4 1 1 1 1 + cos 4π₯ = β« ππ₯ + β« 2cos 2π₯ππ₯ + β« 4 4 4 2 1 1 1 1 = β« ππ₯ + β« 2cos 2π₯ ππ₯ + β« ππ₯ + β« cos 4π₯ ππ₯ 4 4 8 8 1 2π ππ 2π₯ 1 π ππ 4π₯ = (π₯ + ) + (π₯ + )+πΆ 4 2 8 4 1 1 1 1 = π₯ β π ππ 2π₯ + π₯ β π ππ 4π₯ + πΆ 4 4 8 32 3 1 1 = π₯ β π ππ 2π₯ + π ππ 4π₯ + πΆ 8 4 32 Ejercicio 2. β«
β«
π π πππππ
π π = β« π ππ 6 π₯ππ₯ ; ππππ π = β« π ππ 4π₯π ππ 2π₯ππ₯ ; = β«(π ππ 2π₯)2 π ππ 2π₯ππ₯ ; = β«(1 + π‘ππ2 π₯)2 π ππ 2π₯ππ₯ ;
π‘ππ2 π₯ + 1 = π ππ 2 π₯
= β«(π‘ππ4 π₯ + 2π‘ππ2 π₯ + 1) π ππ 2π₯ππ₯ ; = β«(π‘ππ4 π₯ π ππ 2 π₯ + 2π‘ππ2 π₯ π ππ 2π₯ + π ππ 2 π₯ ) ππ₯ ; = β« π‘ππ4 π₯ π ππ 2 π₯ ππ₯ + β« 2π‘ππ2 π₯ π ππ 2 π₯ ππ₯ + β« π ππ 2 π₯ ππ₯ ; = β« π‘ππ4 π₯ π(π‘πππ₯) + 2 β« π‘ππ2 π₯ π(π‘πππ₯) + π‘πππ₯ ; =
1 1 π‘ππ5 π₯ + π‘ππ3 π₯ + tan π₯ + πΆ ; 5 3
Ejercicio 3. β« ππππ ππ π π
β« ππππ ππ π π = β« πππ‘ 2 3π₯(πππ‘ 2 3π₯) ππ₯ ; = β« πππ‘ 2 3π₯(ππ π 2 3π₯ β 1)ππ₯ ; π’ = cot 3π₯ ππ’ = β csc 2 3π₯(3)ππ₯ ππ’ ππ₯ = β 3csc 2 3π₯
= β« cot 2 3 π₯ ππ π 2 3π₯ππ₯ β cot 2 3 π₯ ππ₯ ; = β« cot 2 3 π₯ ππ π 23π₯ππ₯ β β« cot 2 3π₯ ππ₯ ; = β« π’2 csc2 3π₯ (β
ππ’ 3csc2 3π₯
) β β« csc2 3π₯ β 1 ππ₯ ;
1 = β β« π’2 ππ’ β β« csc2 3π₯ + β« ππ₯ 3 =β
1 π’3 cot 3π₯ )+π₯ β (β 3 3 3
=β
π’3 1 + cot 3π₯ + π₯ + πΆ ; 9 3
=β
πππ‘ 3 3π₯ 1 + cot 3π₯ + π₯ + πΆ ; 9 3
Ejercicio 4. β« ππππ π π π
β« ππππ π π π = β«(πππ‘ 2 π₯ + 1)2 (ππ π 2 π₯) ππ₯ ;
π’ = cot π₯ ππ’ = β csc 2 π₯ ππ₯ ππ’ ππ₯ = β csc 2 π₯
= β«(πππ‘ 4 π₯ + 2 cot 2 π₯ + 1)(ππ π 2 π₯)ππ₯ ; = β« cot 4 π₯ ππ π 2 π₯ + 2 cot 2 π₯ππ π 2 π₯ + csc2 π₯ ππ₯ ;
= β« cot 4 π₯ ππ π 2 π₯ππ₯ + β« 2 cot 2 π₯ππ π 2 π₯ππ₯ + β« csc2 π₯ ππ₯ ; = β« π’4 ππ π 2π₯ (β
ππ’ csc2
) + 2 β« u2 π₯ππ π 2 π₯ (β
π₯
= β β« π’4 ππ’ β 2 β« u2 ππ’ + β« csc2 π₯ ππ₯ ; π’5 2π’3 =β β β cot π₯ + πΆ 5 3 πππ‘ 5 π₯ 2πππ‘ 3 π₯ =β β β cot π₯ + πΆ 5 3
ππ’ csc2
) + β« csc2 π₯ ππ₯ ;
π₯
Ejercicio 6. β« ππππ πππππ π π π
β« ππππ πππππ π π π = β« tan4 π₯ π ππ 6 π₯ sec π₯ tan π₯ ππ₯ ; = β«(sec2 π₯ β 1)2 sec6 π₯ π πππ₯ tan π₯ ππ₯ ; = β«(sec4 π₯ β 2 sec2 π₯ + 1) sec6 π₯ π πππ₯ tan π₯ ππ₯ ; = β« (sec10 π₯ β 2 sec8 π₯ + sec6 π₯) π πππ₯ tan π₯ ππ₯ ; = β« (u10 β2 u8 + u6 ) π πππ₯ tan π₯ π’11 2π’9 π’7 = β + +πΆ 11 9 7 π ππ 11π₯ 2π ππ 9 π₯ π ππ 7π₯ = β + +πΆ 11 9 7
ππ’ sec π₯ π‘πππ₯
;
π’ = sec π₯ ππ’ = π πππ₯ tan π₯ ππ₯ ππ’ ππ₯ = sec π₯ π‘πππ₯
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