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!
