la solucion anterior es una forma, ahi te dejo otra: united states of america la sustitucion popular para funciones racionales de seno y coseno z = tan(x/2) x = 2 arctan(z) dx = 2 dz / (z²+one million) sen x = 2 z / (z²+one million) cos x = (z²-one million) / (z²+one million) reemplazando ? [one million/(sen x + cos x ) ] dx = ?[one million/ ( 2 z / (z²+one million) + (z²-one million) / (z²+one million) )] [2 dz / (z²+one million)] simplificando = ?[one million/ ( 2 z + z² - one million) / (z²+one million)] [2 dz / (z²+one million)] = ?[(z²+one million) / ( 2 z + z² - one million)] [2 dz / (z²+one million)] = ?[2 / ( 2 z + z² - one million)] dz completando cuadrados = ?[2 / ( z² + 2z +one million - 2)] dz = ?[2 / (( z+one million)² - (?2)²)] dz para operar mejor una sustitucion adicional u = z+one million du = dz = ?[2 / (u² - (?2)²)] du necessary de tabla ?[one million / (u² - a²)] du = (one million/2a) Ln | (u-a)/u+a)| .... ó ...... (one million/a) arctanh (u/a) luego quedaria asi, usando la primera formula: = 2 (one million/2?2) Ln | (u-?2)/(u+?2)| + ok regresando a z = 2 (one million/2?2) Ln | (z+one million-?2)/(z+one million+?2)| + ok regresando a x = 2 (one million/2?2) Ln | (tan (x/2) + one million-?2)/(tan(x/2) + one million+?2)| + ok = (?2/2) Ln | (tan(x/2) + one million-?2)/(tan(x/2) + one million+?2)| + ok o si prefieres ponerlo en senos y cosenos = (?2/2) Ln | (sen(x/2) + (one million-?2) cos(x/2) )/(sen(x/2) + (one million+?2)cos(x/2) | + ok si deseas pasarlo a x es solo usar las formula de angulo mitad si hubieras usado la otra forma de la necessary de tabla daría: = 2 (one million/?2) arctanh (u/?2) + ok regresando a z = 2 (one million/?2) arctanh [ (z+one million)/?2] + ok regresando a x = 2 (one million/?2) arctanh [ (tan(x/2) +one million) / ?2 ] + ok = ?2 arctanh [ (tan (x/2) +one million) / ?2 ] + ok
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Hola,
(no es una integral simple)
∫ [(cosx + x senx - 1) /(senx - x)²] dx =
apliquemos, en el numerador, la identidad 1 = sen²x + cos²x:
∫ {[cosx + x senx - (sen²x + cos²x)] /(senx - x)²} dx =
∫ [(cosx + x senx - sen²x - cos²x) /(senx - x)²] dx =
agrupemos los términos como:
∫ {[(cosx - cos²x) + (x senx - sen²x)] /(senx - x)²} dx =
saquemos los factores comunes de cada grupo:
∫ {[cosx (1 - cosx) + senx (x - senx)] /(senx - x)²} dx =
distribuyamos y simplifiquemos:
∫ { {[cosx (1 - cosx)] /(senx - x)²} + {[senx (x - senx)] /(senx - x)²} } dx =
∫ { {[cosx (1 - cosx)] /(senx - x)²} - {[senx (senx - x)] /(senx - x)²} dx =
∫ { {[cosx (1 - cosx)] /(senx - x)²} - [senx /(senx - x)]} dx =
(partiendo en dos integrales)
∫ {[cosx (1 - cosx)] /(senx - x)²} dx - ∫ [senx /(senx - x)]} dx (#)
escribamos la primera integral como:
∫ cosx [(senx - x)‾ ² (1 - cosx) dx] =
integremos por partes, poniendo:
cosx = u → - senx dx = du
(senx - x)‾ ² (1 - cosx) dx = dv
(cambiando los signos para el diferencial de de senx - x)
- (senx - x)‾ ² (cosx - 1) dx = dv
- (senx - x)‾ ² d(senx - x) = dv
- [1/(- 2+1)] (senx - x)‾ ² ⁺ ¹ = v
- [1/(- 1)] (senx - x)‾ ¹ = v
1 /(senx - x) = v
obteniendo:
∫ u dv = v u - ∫ v du
∫ cosx [(senx - x)‾ ² (1 - cosx) dx] = [1 /(senx - x)] cosx - ∫ [1 /(senx - x)] (- senx) dx =
[cosx /(senx - x)] + ∫ [senx /(senx - x)] dx
luego la expresión anterior (#) se vuelve:
[cosx /(senx - x)] + ∫ [senx /(senx - x)] dx - ∫ [senx /(senx - x)]} dx =
borrando los términos opuestos concluimos con:
[cosx /(senx - x)] + C
espero que sea de ayuda
¡Saludos!
la solucion anterior es una forma, ahi te dejo otra: united states of america la sustitucion popular para funciones racionales de seno y coseno z = tan(x/2) x = 2 arctan(z) dx = 2 dz / (z²+one million) sen x = 2 z / (z²+one million) cos x = (z²-one million) / (z²+one million) reemplazando ? [one million/(sen x + cos x ) ] dx = ?[one million/ ( 2 z / (z²+one million) + (z²-one million) / (z²+one million) )] [2 dz / (z²+one million)] simplificando = ?[one million/ ( 2 z + z² - one million) / (z²+one million)] [2 dz / (z²+one million)] = ?[(z²+one million) / ( 2 z + z² - one million)] [2 dz / (z²+one million)] = ?[2 / ( 2 z + z² - one million)] dz completando cuadrados = ?[2 / ( z² + 2z +one million - 2)] dz = ?[2 / (( z+one million)² - (?2)²)] dz para operar mejor una sustitucion adicional u = z+one million du = dz = ?[2 / (u² - (?2)²)] du necessary de tabla ?[one million / (u² - a²)] du = (one million/2a) Ln | (u-a)/u+a)| .... ó ...... (one million/a) arctanh (u/a) luego quedaria asi, usando la primera formula: = 2 (one million/2?2) Ln | (u-?2)/(u+?2)| + ok regresando a z = 2 (one million/2?2) Ln | (z+one million-?2)/(z+one million+?2)| + ok regresando a x = 2 (one million/2?2) Ln | (tan (x/2) + one million-?2)/(tan(x/2) + one million+?2)| + ok = (?2/2) Ln | (tan(x/2) + one million-?2)/(tan(x/2) + one million+?2)| + ok o si prefieres ponerlo en senos y cosenos = (?2/2) Ln | (sen(x/2) + (one million-?2) cos(x/2) )/(sen(x/2) + (one million+?2)cos(x/2) | + ok si deseas pasarlo a x es solo usar las formula de angulo mitad si hubieras usado la otra forma de la necessary de tabla daría: = 2 (one million/?2) arctanh (u/?2) + ok regresando a z = 2 (one million/?2) arctanh [ (z+one million)/?2] + ok regresando a x = 2 (one million/?2) arctanh [ (tan(x/2) +one million) / ?2 ] + ok = ?2 arctanh [ (tan (x/2) +one million) / ?2 ] + ok