Du U U U U: )] ( Cot ) ( ).[ Csc(
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Segunda parte (punto 5 al 8) El conjunto de todas las antiderivadas de f(x) se llama integral indefinida de f respecto a x, y se denota por el sΓmbolo β« π(π)π
π = π(π) + πͺ. Resolver aplicando las propiedades bΓ‘sicas, las siguientes integrales: 5.
u (u 2 ο 16) ο² u 2 ο« 4u du β«
π’(π’ β 4)(π’ + 4) π’(π’ β 4) ππ’ = β« ππ’ = β«(π’ β 4)ππ’ π’(π’ + 4) π’ = β« π’ ππ’ β β« 4ππ’
=
π’1+1 β 4π’ + πΆ 1+1
=
6.
π’2 β 4π’ + πΆ 2
ο² csc( x).[ sen ( x) ο« cot ( x)] dx =β«
π ππ¨π¬ π π π¬π’π§π π + ππ¨π¬ π [π¬π’π§ π + ] π π = β« [ ] π π π¬π’π§ π π¬π’π§ π π¬π’π§ π π¬π’π§ π
π¬π’π§π π + ππ¨π¬ π = β«[ ] π π π¬π’π§π π = β« [π +
ππ¨π¬ π ] π π π¬π’π§π π
= β«[π]π π + β« [
ππ¨π¬ π ] π π π¬π’π§π π
Usando la sustituciΓ³n π’ = sin π₯ ππ’ = πππ π₯ ππ₯ ππ₯ = β«[ Resolviendo la integral:
ππ’ cos π₯
cos π₯ ππ’ ππ’ 1 ] =β« 2 =β 2 sin π₯ cos π₯ π’ π’
= β«[π]π π + β« [
ππ¨π¬ π π ] π π = π β +πͺ π π¬π’π§ π ππππ
= π β πππ π + πͺ
2x ο« 3 ο² x ο« 2 dx
7.
π’π ππππ ππ π π’π π‘ππ‘π’ππππ: π’ = π₯ + 2 π₯ =π’β2 β«
2(π’ β 2) + 3 2π’ β 4 + 3 2π’ β 1 ππ’ = β« ππ’ = β« ππ’ π’ π’ π’ 1 1 = β« (2 β ) ππ’ = β« 2ππ’ β β« ππ’ π’ π’ = 2π’ β ln π’ + πΆ
= π(π + π) β ππ(π + π) + πͺ
dx ο² 1 ο« sen( x)
8.
ππ₯
1βsin(π₯)
β« 1+sin(π₯) 1βsin(π₯) = β«
1βsin(π₯) cos2(π₯)
1
sin(π₯)
ππ₯ = β« cos2 π₯ ππ₯ β β« cos2 π₯ ππ₯
1
β« cos2 π₯ ππ₯ = tan(π₯) + πΆ
sin(π₯)
β« cos2 π₯ ππ₯ Usando la sustituciΓ³n: π’ = cos π₯ ππ’ = βπ πππ₯ ππ₯ ππ₯ = β sin(π₯)
β« cos2 π₯ ππ₯ = β β«
sin π₯ ππ’ π’2
π πππ₯
ππ’ π πππ₯
ππ’
π’β2+1
1
= β« π’2 ππ’ = β β2+1 = π’β1 = πππ π₯ = sec π₯
La soluciΓ³n de la integral serΓ‘ entonces:
β«
ππ₯ ππ₯ = tan(π₯) + sec(π₯) + πΆ 1 + sin(π₯)
u (u 2 ο 16) ο² u 2 ο« 4u du β«
π’(π’ β 4)(π’ + 4) π’(π’ β 4) ππ’ = β« ππ’ = β«(π’ β 4)ππ’ π’(π’ + 4) π’ = β« π’ ππ’ β β« 4ππ’
=
π’1+1 β 4π’ + πΆ 1+1
=
6.
π’2 β 4π’ + πΆ 2
ο² csc( x).[ sen ( x) ο« cot ( x)] dx =β«
π ππ¨π¬ π π π¬π’π§π π + ππ¨π¬ π [π¬π’π§ π + ] π π = β« [ ] π π π¬π’π§ π π¬π’π§ π π¬π’π§ π π¬π’π§ π
π¬π’π§π π + ππ¨π¬ π = β«[ ] π π π¬π’π§π π = β« [π +
ππ¨π¬ π ] π π π¬π’π§π π
= β«[π]π π + β« [
ππ¨π¬ π ] π π π¬π’π§π π
Usando la sustituciΓ³n π’ = sin π₯ ππ’ = πππ π₯ ππ₯ ππ₯ = β«[ Resolviendo la integral:
ππ’ cos π₯
cos π₯ ππ’ ππ’ 1 ] =β« 2 =β 2 sin π₯ cos π₯ π’ π’
= β«[π]π π + β« [
ππ¨π¬ π π ] π π = π β +πͺ π π¬π’π§ π ππππ
= π β πππ π + πͺ
2x ο« 3 ο² x ο« 2 dx
7.
π’π ππππ ππ π π’π π‘ππ‘π’ππππ: π’ = π₯ + 2 π₯ =π’β2 β«
2(π’ β 2) + 3 2π’ β 4 + 3 2π’ β 1 ππ’ = β« ππ’ = β« ππ’ π’ π’ π’ 1 1 = β« (2 β ) ππ’ = β« 2ππ’ β β« ππ’ π’ π’ = 2π’ β ln π’ + πΆ
= π(π + π) β ππ(π + π) + πͺ
dx ο² 1 ο« sen( x)
8.
ππ₯
1βsin(π₯)
β« 1+sin(π₯) 1βsin(π₯) = β«
1βsin(π₯) cos2(π₯)
1
sin(π₯)
ππ₯ = β« cos2 π₯ ππ₯ β β« cos2 π₯ ππ₯
1
β« cos2 π₯ ππ₯ = tan(π₯) + πΆ
sin(π₯)
β« cos2 π₯ ππ₯ Usando la sustituciΓ³n: π’ = cos π₯ ππ’ = βπ πππ₯ ππ₯ ππ₯ = β sin(π₯)
β« cos2 π₯ ππ₯ = β β«
sin π₯ ππ’ π’2
π πππ₯
ππ’ π πππ₯
ππ’
π’β2+1
1
= β« π’2 ππ’ = β β2+1 = π’β1 = πππ π₯ = sec π₯
La soluciΓ³n de la integral serΓ‘ entonces:
β«
ππ₯ ππ₯ = tan(π₯) + sec(π₯) + πΆ 1 + sin(π₯)
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