Another Integration Question :S

[Redoctober]

Homework Statement

∫(sin(x-2))^(1/3)

The Attempt at a Solution

I reached until -∫1/((1-u^2)^1/3).du where u = sin(x)
Then i got stuck as i tried to integrate to inverse function but it doesn't work as that is for ∫1/((a-x^2)^1/2)

another try was to separate the sin(x-2)

where i reached until

∫(1-(cos(x-2))^2)*(sin(x-2))^(5/3).dx
then using u= cos(x-2) u get -du=sin(x-2) . but this doesn't work too -.- !

Help

 

[mtayab1994]

Do you know this or no?

 

[Redoctober]

Do you know this or no?

View attachment 42204

Yes , this is the function of integration .
If u mean ∫(sin(x-2))^(1/3).dx = 3/4*cos(x-2)^(4/3) + C . it is wrong :S

 

[SammyS]

Homework Statement

∫(sin(x-2))^(1/3)

The Attempt at a Solution

I reached until -∫1/((1-u^2)^1/3).du where u = sin(x)
Then i got stuck as i tried to integrate to inverse function but it doesn't work as that is for ∫1/((a-x^2)^1/2)

another try was to separate the sin(x-2)

where i reached until

∫(1-(cos(x-2))^2)*(sin(x-2))^(5/3).dx
then using u= cos(x-2) u get -du=sin(x-2) . but this doesn't work too -.- !

Help

Hello Redoctober.

WolframAlpha leads me to believe that this can't be expressed in terms of elementary functions.

You might try a power series method.

By The Way: You should be posting these integration questions in the Calculus and Beyond section.

 

[Redoctober]

Hello Redoctober.

WolframAlpha leads me to believe that this can't be expressed in terms of elementary functions.

You might try a power series method.

By The Way: You should be posting these integration questions in the Calculus and Beyond section.

Oh k thanks :) . I think i ll keep it for later , as we didn't start yet with the power series method yet in the Calculus 1 course .

K i ll post in the Calculus and Beyond :) .