Hola,
iniciemos con una sustitución; sea:
arcsen(√x) = t
√x = sen t
x = sen²t
dx = 2sen t cos t dt
substituyendo, obtenemos:
∫ arcsen(√x) dx = ∫ t (2sen t cos t) dt =
(aplicando la fórmula del ángulo doble sen(2t) = 2sen t cos t)
∫ t sen(2t) dt =
integremos por partes, poniendo:
t = u → dt = du
sen(2t) dt = dv → (1/2)[- cos(2t)] = v
obteniendo:
∫ u dv = v u - ∫ v du
∫ t sen(2t) dt = (1/2)[- cos(2t)] t - ∫ (1/2)[- cos(2t)] dt =
- (1/2)t cos(2t) + (1/2) ∫ cos(2t) dt =
- (1/2)t cos(2t) + (1/2) (1/2)sen(2t) + C =
(aplicando las fórmulas del ángulo doble cos(2t) = cos²t - sen²t y sen(2t) = 2sen t cos t)
- (1/2)t (cos²t - sen²t) + (1/2) (1/2)2sen t cos t + C =
- (1/2)t cos²t + (1/2)t sen²t + (1/2)sen t cos t + C
recordemos que:
t = arcsen(√x)
sen t = √x
por tanto:
cos t = √(1 - sen²t) = √[1 - (√x)²] = √(1 - x)
luego, substituyendo:
- (1/2)t cos²t + (1/2)t sen²t + (1/2)sen t cos t + C = - (1/2)arcsen(√x) [√(1 - x)]² + (1/2)arcsen(√x) (√x)² + (1/2)(√x) √(1 - x) + C =
- (1/2)arcsen(√x) (1 - x) + (1/2)arcsen(√x) x + (1/2) √[x (1 - x)] + C =
- (1/2)(1 - x) arcsen(√x) + (1/2)x arcsen(√x) + (1/2)√(x - x²) + C =
(desarrollando)
- (1/2)arcsen(√x) + (1/2)x arcsen(√x) + (1/2)x arcsen(√x) + (1/2)√(x - x²) + C =
- (1/2)arcsen(√x) + 2(1/2)x arcsen(√x) + (1/2)√(x - x²) + C
pues la respuesta es:
∫ arcsen(√x) dx = - (1/2)arcsen(√x) + x arcsen(√x) + (1/2)√(x - x²) + C
espero haber sido de ayuda
¡Saludos!
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Answers & Comments
Verified answer
Hola,
iniciemos con una sustitución; sea:
arcsen(√x) = t
√x = sen t
x = sen²t
dx = 2sen t cos t dt
substituyendo, obtenemos:
∫ arcsen(√x) dx = ∫ t (2sen t cos t) dt =
(aplicando la fórmula del ángulo doble sen(2t) = 2sen t cos t)
∫ t sen(2t) dt =
integremos por partes, poniendo:
t = u → dt = du
sen(2t) dt = dv → (1/2)[- cos(2t)] = v
obteniendo:
∫ u dv = v u - ∫ v du
∫ t sen(2t) dt = (1/2)[- cos(2t)] t - ∫ (1/2)[- cos(2t)] dt =
- (1/2)t cos(2t) + (1/2) ∫ cos(2t) dt =
- (1/2)t cos(2t) + (1/2) (1/2)sen(2t) + C =
(aplicando las fórmulas del ángulo doble cos(2t) = cos²t - sen²t y sen(2t) = 2sen t cos t)
- (1/2)t (cos²t - sen²t) + (1/2) (1/2)2sen t cos t + C =
- (1/2)t cos²t + (1/2)t sen²t + (1/2)sen t cos t + C
recordemos que:
t = arcsen(√x)
sen t = √x
por tanto:
cos t = √(1 - sen²t) = √[1 - (√x)²] = √(1 - x)
luego, substituyendo:
- (1/2)t cos²t + (1/2)t sen²t + (1/2)sen t cos t + C = - (1/2)arcsen(√x) [√(1 - x)]² + (1/2)arcsen(√x) (√x)² + (1/2)(√x) √(1 - x) + C =
- (1/2)arcsen(√x) (1 - x) + (1/2)arcsen(√x) x + (1/2) √[x (1 - x)] + C =
- (1/2)(1 - x) arcsen(√x) + (1/2)x arcsen(√x) + (1/2)√(x - x²) + C =
(desarrollando)
- (1/2)arcsen(√x) + (1/2)x arcsen(√x) + (1/2)x arcsen(√x) + (1/2)√(x - x²) + C =
- (1/2)arcsen(√x) + 2(1/2)x arcsen(√x) + (1/2)√(x - x²) + C
pues la respuesta es:
∫ arcsen(√x) dx = - (1/2)arcsen(√x) + x arcsen(√x) + (1/2)√(x - x²) + C
espero haber sido de ayuda
¡Saludos!