Respuestas

  • Usuario de Brainly
2012-04-18T01:49:03+02:00

Como sugerido, pongamos:

tan(x/2) = z

x/2 = arctan z

x = 2arctan z

dx = 2[1/(1 + z²)] dz

además, recordemos las identidades:

senx = 2tan(x/2) /[1 + tan²(x/2)] = 2z/(1 + z²)

cosx = [1 - tan²(x/2)] /[1 + tan²(x/2)] = (1 - z²)/(1 + z²)

tanx = senx/cosx = [2z /(1 + z²)] /[(1 - z²)/(1 + z²)] = [2z/(1 + z²)][(1 + z²)/(1 - z²)] = 
2z/(1 - z²)

luego, substituyendo:

dx /(tanx + senx) = 2[1 /(1 + z²)] dz /{[2z /(1 - z²)] + [2z /(1 + z²)]} =

[2 /(1 + z²)] dz /{[2z(1 + z²) + 2z(1 - z²)] /[(1 - z²)(1 + z²)]} =

[2 /(1 + z²)] dz /{(2z + 2z³ + 2z - 2z³) /[(1 - z²)(1 + z²)]} =

[2 /(1 + z²)] dz /{4z /[(1 - z²)(1 + z²)]} =

[2 /(1 + z²)] {[(1 - z²)(1 + z²)] /(4z)} dz =

(simplificando y llevando fuera las constantes)

(1/2) [(1 - z²) /z] dz =

partamos la integral en:

(1/2) (1/z) dz - (1/2) (z²/z) dz =

(1/2) ln | z | - (1/2) z dz =

(1/2) ln | z | - (1/2) [1/(1+1)] z^(1+1) + C =

(1/2) ln | z | - (1/2)(1/2)z² + C =

(1/2) ln | z | - (1/4)z² + C 

en fin substituyamos de nuevo z = tan(x/2), concluyendo con:


dx /(tanx + senx) = (1/2) ln |tan(x/2)| - (1/4)tan²(x/2) + C 



espero haber sido de ayuda..
¡Saludos!