Hi Everyone,
Please help me with this integration:
This is what I've tried so far:
when I substitute
and
I got:
the new boundaries is my main problem here, it will give me a value of 0.
However, when I check my answer, it should be
Hi Everyone,
Please help me with this integration:
This is what I've tried so far:
when I substitute
and
I got:
the new boundaries is my main problem here, it will give me a value of 0.
However, when I check my answer, it should be
The problem is that when you replace the limits, you're changing the definition for the path of integration. Usually just substituting the limits works: if , then "s goes from 4 to 12" and "t goes from 1 to 3" mean the same thing.
In your case, however, since , as x goes from 0 to , z goes from 1 to 2 (when ) and then back to 1 again. That's not necessarily a problem, but your equation for x is different for the two ranges.
For x between 0 and , we want , but for x between and , we want (since the inverse sine function is defined to be between and ). When you plug into the integrand, you get the same thing, but when you plug into dx, you get a negative sign on one but not the other.
Once you see the problem, you work around it: since the integrand is symmetric around ,
.
- Hollywood
Hi Hollywood,
Thanks for answering....
Can I say:
whenever the curve is oscillating along x-axis (or it is a one-to-many function) when we integrate it by substitution, we should break it into few parts (local minimum to local maximum / local maximum to local minimum) to solve it?
No, the substitution method can still work (and usually does work, in my experience) even if the substitution function is not one-to-one. For example,
Substituting , gives
but as x goes from -1 to 2, u goes from 1 to 0 and then back to 1 on its way to 4.
- Hollywood
By the way, in my previous post the integrand should be , not . Sorry about that.