1. ## Double integrals problem

Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
$\int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx dy$
...and I'm told to solve it by changing the order of integration. I get:
$\int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$
I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate $2x^5y\cos{(xy^2)}$ with respect to y, taking x as a constant.

Can anyone help?

2. Originally Posted by chella182
Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
$\int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx dy$
...and I'm told to solve it by changing the order of integration. I get:
$\int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$
I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate $2x^5y\cos{(xy^2)}$ with respect to y, taking x as a constant.

Can anyone help?
I didn't check your change of order of inegration but use this sub to evaluate the integral

$u=xy^2 \implies du=2xydy$

and we get

$\int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$

$\int_{0}^{1}\int x^4\cos(u)dudx$

3. Originally Posted by chella182
Well, my problem isn't with the double integral bit, it's just with the integration once I've switched the limits et cetera. The original problem is:
$\int_{0}^{1}\int_{\sqrt{y}}^{1}2x^5y\cos{(xy^2)}dx dy$
...and I'm told to solve it by changing the order of integration. I get:
$\int_{0}^{1}\int_{0}^{x^2}2x^5y\cos{(xy^2)}dydx$
I know this forum's not here to check my work, but whether my new limits are correct or not, I still don't know how to integrate $2x^5y\cos{(xy^2)}$ with respect to y, taking x as a constant.

Can anyone help?
Your reordering of the integration is correct. Proceeding from there, let $u=y^2$. Thus $du = 2y\,dy$

You now have $\int_0^1\int_0^{x^2}x^5\cos(xu)\,du = \frac{1}{x}\cdot x^5\sin(xu) = \left[x^4\sin(xy^2)\right]_0^{x^2} = x^4\sin(x^5)$

So we have $\int_0^1 x^4\sin(x^5)\,dx$

Let $u=x^5$ and $du=5x^4$. Now you have:

$\frac{1}{5}\int_0^1\sin(u)\,du = -\frac{1}{5}\cos(u) = \left[-\frac{1}{5}\cos(x^5)\right]_0^1 = \boxed{\frac{1}{5}-\frac{1}{5}\cos(1)}$

4. I totally don't understand the second line of your working, sorry where does the $2y$ disappear to? And the $x^4$ in the second integral? And I don't get where the $\frac{1}{x}$ has come from

5. $2y$ is the derivative of $y^{2}$. All he did was a u-substitution.

6. No, I know that, where does it disappear to though?

7. Into the $du$. It never "disappears", it's just part of his new variables $u$ and $du$. And the $\frac{1}{x}$ is part of the integration of $cos(xy)$.

The antiderivative of $cos(xy)$ where $x$ is treated as a variable is $\frac{sin(xy)}{x}$.

8. So because the $2y$ appears in the $du$ it's not in the problem anymore? :S I'm well confused.

I get where the $\frac{1}{x}$ comes from now though, cheers.

9. It's just u-substitution.

Just replace the $y^{2}$ with a $u$ and $2ydy$ with a $du$. Don't let the fact that it's a double integral throw you off. It's just a straightforward u-substitution.

10. I know I just... don't get it :S urgh. I'll go see my lecturer tomorrow if I get chance.

11. Let's look at a single integral:

$\int 2ycos(y^{2})\,\,dy$

Let $u=y^{2}$ and $du=2ydy$.

Now just substitute:

$\int cos(u)\,\,du$

Now just integrate:

$\int cos(u)\,\,du = sin(u)$

Now just resubstitute for $u=y^{2}$:

$sin(y^{2})$.

Take the derivative to clarify:

$\frac{d}{dy}sin(y^{2})=cos(y^{2})\cdot \frac{d}{dy}y^{2} = 2ycos(y^{2})$

All you're doing is substituting one type of variable for another. The actual function isn't changed, just the way it's being represented.

12. Ohhh right, I get it. All you needed to do to explain was say $dy=\frac{du}{2y}$ so the $2y$'s cancel

13. It hasn't disappeared. $du=2ydy$

I'm just representing $2ydy$ as $du$ so I can have an easier integral to evaluate.

And yes, you can think of it that way if you want. Whatever clicks with you.

14. It has cancelled out therefore it's disappeared in my mind. I do understand the whole thing, all I didn't get was why that disappeared 'cause no one was telling me where it'd gone. Although this does seem rather more complicated than examples we've done in lectures :S