integrales

1. ∫ sen^3 x dx =

si sen²x + cos²x = 1
sen²x = 1 - cos²x

∫ sen^3 x dx = ∫ senx . sen²x dx = ∫ senx . (1-cos²x) dx

= ∫ (senx - senx cos²x) dx

= ∫ sen dx - ∫ senx cos²x dx

Resolviendo la segunda integral por sustitución, es

cos x = u --> -sen x dx = du --> senx dx = - du

=∫ sen dx - ∫ u² (-1) du

= - cosx + (u^3)/3 + c

= - cosx + (cos^3)/3 + c

2. ∫ cos^3 x dx =si sen²x + cos²x = 1
cos²x = 1 - sen²x

∫ cos^3 x dx = ∫ cos x cos²x dx

∫ cos x cos²x dx

∫ cos x (1 - sen²x)dx

∫ cos x dx -∫cos x sen²xdx

Sen x = u --> cos x dx = du --> senx dx = du

∫ cos x dx -∫u²xdu

=sen x –u^3/3+C

= sen x –sen^3 x/3 + C

3. ∫ cos ^2 x sen^3 x dx

si sen²x + cos²x = 1
sen²x = 1 - cos²x

∫ cos^2 x sen^3 x dx = ∫ cos^2 x senx . sen²x dx = ∫ cos ^2senx . (1-cos²x) dx

= ∫ (cos ^2 senx - senx cos^4x) dx

= ∫ cos ^2 sen dx - ∫ senx cos^4x dx

Resolviendo la segunda integral por sustitución, es

cos x = u --> -sen x dx = du --> senx dx = - du

= -∫ u^2 du + ∫ u^4 du

=-u^3/3+u^5/5 + C
= -cos^3 x/3+cos ^5 x/5 +c

4. ∫ sen^5 x cos^4 x dx =

si sen²x + cos²x = 1
sen²x = 1 - cos²x

= ∫sen^5 x cos^4 x dx = ∫senx . (sen²x ) ^2cos^4 x dx =∫cos ^4 senx . (1-cos²x)^2dx

= ∫ cos ^4 x senx . (1-2cos²x+cos^4 x) dx

= ∫ cos ^4 x senx .dx - 2∫cos^6 x sen x dx + ∫ cos^8 x sen x dx

cos x = u --> -sen x dx = du --> senx dx = - du

= -∫u^4 du + ∫u^6 du- ∫ u^8 du=

=-u^5/5+ 2 u^7/7 - u^9/9 + c

=- cos ^5 x/5 + 2 cos ^7 x/7- cos ^9 x/9+ c

5. ∫ cos(4x) cos(7x) dx =

cosα cosβ = (1/2) [cos(α + β) + cos(α - β)]

∫cos(4x) cos(7x) dx = ∫ (1/2) {cos[(4x) + (7x)] + cos[(4x) - (7x)]} dx =

= ∫ (1/2) [cos(11x) + cos(- 3x)] dx =

cos(- α) = cosα
= ∫ (1/2) [cos(11x) + cos(3x)] dx = (1/2) ∫ cos(11x) dx + (1/2) ∫ cos(3x) dx =

= (1/2) (1/11)sin(11x) + (1/2) (1/3)sin(3x) + c =

= ∫ cos(4x) cos(7x) dx = (1/22) sin(11x) + (1/6) sin(3x) + c

6. ∫ sen(3x) cos(5x) dx

senα cosβ = (1/2) [sen(α + β) + sen(α -β)]= (1/2) [sen(3x+5x) + sen(3x – 5x)]=

= (1/2) [sen(8x) + sen(– 2x)]

= ∫ (1/2)[sen(8x) + sen(-2x)] dx= (1/2) ∫ sen(8x)dx + (1/2)∫sen(-2x)dx

U= 8x du=8 dx du/8=dx

V=-2x dv= -2dx -dv/2= dx

=1/2* 1/8 ∫sen u du + 1/2 * (-1/2) ∫sen v dv

= -1/16 cos u -1/4 cos v + c

= -1/16 cos (8x) + 1/4 cos (-2x) + c

7. ∫ tg^4x.dx = ∫ tg²x.(sec²x - 1).dx

∫ tg²x.(sec²x - 1).dx = ∫tg²x.sec²x .dx - ∫ tg²x dx

∫ tg²x.sec²x .dx - ∫ tg²x dx = ∫ tg²x.sec²x .dx - ∫ (sec²x - 1)dx
∫ tg²x.sec²x .dx - ∫ (sec²x - 1)dx = ∫ tg²x.sec²x .dx - tgx + x

Solo queda resolver la integral:

∫ tg²x.sec²x .dx 

Por sustitución

y = tgx

dy = sec²x.dx -------> dx = dy /sec²x

∫ tg²x.sec²x .dx = ∫ y².sec²x dy/sec²x

∫ y².sec²x dy/sec²x = ∫ y².dy = y³/3 +C

Restituyendo

∫ tg²x.sec²x.dx = tg³x/3 + C

Finalmente

∫ tg^4x.dx = tg³x/3 - tgx + x + C

8. ∫cotg^6 x dx 

cotg²(x) = cosec²(x) – 1

d(cotg(x)) = - cosec²(x) = -1 -cotg²(x)

∫cotg^6 x dx 

= ∫cotg² x· cotg^4 x dx 

= ∫(cosec² x - 1) · cotg^4 x dx 

= ∫cosec² x·cotg^4 x·dx - ∫cotg^4 x dx (1)

resuelvo la primera integral directamente, o con el cambio de variable 

t=cotg(x) ==> dt=-cosec²(x)∫cosec²(x)·cotg^4(x)·dx = -·∫t^4·dt = -(1/5)t^5

(2) ∫cosec²(2x)·cotg^4(2x)·dx = -(1/5)·cotg^5 ( x)

la segunda integral de (1) se separa como antes

∫cotg^4(x) dx = ∫cotg²(x)·∫cotg²(x)·dx

=∫[cosec²(x)-1]·cotg²(x)·dx

=∫cosec²(x)·cotg²(x)·dx - ∫cotg²(x) dx (3)

la primera se resuelve igual que la anterior

∫cosec²(x)·cotg²(x)·dx = dx = -·∫t^2·dt = -(1/3)t^3= -(1/3)·cotg³(x)

t=cotg(x) ==>dt=-cosec²(x)

la segunda es inmediata

= ∫cotg²(x) dx = ∫ (csc²(x) - 1 ) dx 

= ∫[csc²(2x) dx -∫dx 
= cotg(x) + x

la expresión (3) queda como

∫cotg^4(x) dx = -(1/3)·cotg³(x) +·cotg(x) + x

juntando este resultado y (2) en (1)

∫cotg^6(x) dx = -(1/5)·cotg^5(x) + (1/3)·cotg³(x) - ·cotg(x) – x +c

9. ∫ tg²x sec³x dx=

tg²x=sec²x – 1

= ∫ (sec²x – 1 ) sec³x dx=

= ∫...