Hola Alex! Sí, también puedes hacerlo así! Te muestro como sería el principio, pero si tienes problemas puedo escribirte el final si quieres: Integral de x*cos(ln(x)) dx = Sustitución: t = ln(x) ==> e^t = x dt = 1/x dx ==> x dt = dx ==> e^t dt = dx = Integral de (e^t)cos(t) (e^t)dt = = Integral de (e^2t)cos(t) dt = = ...
Hi Pablo, it becomes very long because the integral of x*cos(ln(x)) becomes the integral of x*sin(ln(x)) and it is like we are "running in circles".. Integral of x*cos(ln(x)) dx = Parts: Integral of u dv = uv - Integral v du u = x ==> du = dx dv = cos(ln(x)) dx ==> v = th-cam.com/video/kHTpZFncmU8/w-d-xo.html = (x/2)[cos(ln(x)) + sin(ln(x))] = x*(x/2)[cos(ln(x)) + sin(ln(x))] - Integral of (x/2)[cos(ln(x)) + sin(ln(x))] dx = = (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*cos(ln(x)) - (1/2)Integral of x*sin(ln(x)) dx ==> Integral of x*cos(ln(x)) dx = = (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*cos(ln(x)) - (1/2)Integral of x*sin(ln(x)) dx ==> Integral of x*cos(ln(x)) dx + (1/2)Integral of x*cos(ln(x)) = = (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*sin(ln(x)) dx ==> (3/2)Integral of x*cos(ln(x)) = = (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*sin(ln(x)) dx
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Al final... ¿No faltó una X² a lado del sen (In (x))?
Totalmente cierto! Gracias por avisar y espero que no te haya causado mucha molestia...
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Profesor y tambien podria haber hecho una sustitucion de u=ln(x) ,e^u=x,dx=e^udu??? y despues aplicar por partes??
Hola Alex! Sí, también puedes hacerlo así! Te muestro como sería el principio, pero si tienes problemas puedo escribirte el final si quieres:
Integral de x*cos(ln(x)) dx =
Sustitución:
t = ln(x) ==> e^t = x
dt = 1/x dx ==> x dt = dx ==> e^t dt = dx
= Integral de (e^t)cos(t) (e^t)dt =
= Integral de (e^2t)cos(t) dt =
= ...
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What if u---- x dv--- cos(ln(x))dx???
Hi Pablo, it becomes very long because the integral of x*cos(ln(x)) becomes the integral of x*sin(ln(x)) and it is like we are "running in circles"..
Integral of x*cos(ln(x)) dx =
Parts: Integral of u dv = uv - Integral v du
u = x ==> du = dx
dv = cos(ln(x)) dx ==> v = th-cam.com/video/kHTpZFncmU8/w-d-xo.html = (x/2)[cos(ln(x)) + sin(ln(x))]
= x*(x/2)[cos(ln(x)) + sin(ln(x))] - Integral of (x/2)[cos(ln(x)) + sin(ln(x))] dx =
= (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*cos(ln(x)) - (1/2)Integral of x*sin(ln(x)) dx
==>
Integral of x*cos(ln(x)) dx =
= (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*cos(ln(x)) - (1/2)Integral of x*sin(ln(x)) dx
==>
Integral of x*cos(ln(x)) dx + (1/2)Integral of x*cos(ln(x)) =
= (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*sin(ln(x)) dx
==>
(3/2)Integral of x*cos(ln(x)) =
= (x^2/2)[cos(ln(x)) + sin(ln(x))] - (1/2)Integral of x*sin(ln(x)) dx
at the end because I am multiplying 5/4 if 5/4 is multiplying the integral, should I not divide?