Double integral of ((x^3)+1)^(1/2)

[troyofyort]

So I have to evaluate the integral from y=0 to y=1 of(the integral from x=(y^(1/2)) to x=1 of ((x^3)+1)^(1/2)dx)dy.

I've substituted the ((x^3)+1) with sec^2(u) since I used tan^2(u)=x^3. I'm wondering if this is the correct (or even a good) manner of solving this because I'm ending up with a very difficult equation to integrate anyways with odd bounds?

 

[jedishrfu]

I don't think you can use the tan^2(u) = x^3 substitution. I've seen people use tan^2(u) = x^2 substitution where the powers are the same.

Isn't there some other substitution you've studied with cubic powers?

Perhaps you could factor x^3 + 1?

 

[troyofyort]

Do you mean factor x^3+1 into (x+1)(x^2-x+1)?
Unfortunately I'm not well versed in trig sub so I'm lost on substitution with cubic powers so I'm currently looking it up.

I'm also thinking it would be easier to solve if I changed the solving for integral of f(x,y)dx first to solving integral of f(x,y)dy first

 

[SammyS]

So I have to evaluate the integral from y=0 to y=1 of(the integral from x=(y^(1/2)) to x=1 of ((x^3)+1)^(1/2)dx)dy.

I've substituted the ((x^3)+1) with sec^2(u) since I used tan^2(u)=x^3. I'm wondering if this is the correct (or even a good) manner of solving this because I'm ending up with a very difficult equation to integrate anyways with odd bounds?

Try changing the order of integration.