For the second...
.
Now let so that and the integral becomes
.
Hi,
I'm studying for a test and I'm doing some integral exercises. I manage to solve a lot of them, but I'm always having difficulty with integrals that have square roots.
I have 4 equations, but I don't want the actual answer to any of them. I would just like a push in the right direction for solving them.
Thanks for any help!
For 1, use parts, integrating root x and differentiating the other, then make a u-sub of the root which is in the denominator of the subsequent integrand.
Just in case a picture helps...
... where (key in spoiler) ...
Spoiler:
Spoiler:
Third one, you can substitute for , though I'm not certain you need to.
Number four is very similar to just the latter part of number one.
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
You're right except for the first coefficient (see my first pic), and if I'm not wrong then the last integral to do is...
See my second spoiler for more detail about this.
Just in case a picture helps with number 3...
... where (key in spoiler) ...
Spoiler:
And the rest...
Spoiler:
_________________________________________
Don't integrate - balloontegrate!
Balloon Calculus; standard integrals, derivatives and methods
Balloon Calculus Drawing with LaTeX and Asymptote!
The substitution to use here is u (or s, or my dashed balloon) equals... because that will do as the inner function of a chain rule differentiation that we need to work backwards through.
I put this...
in the spoiler - not sure if you saw that (I did edit it in a bit later). It doesn't contain any differentials, but hopefully you can see that we've expressed x (your v) in terms of u (my dashed balloon). That's how we know that the function to integrate with respect to whatever we're calling the inner function (u or the dashed balloon or s) is
Hello, superdl!
Here's the 4th one . . .
If there is a linear function under the radical,
. . let equal the entire radical.
Let
. . . . . . .
At this point, you know how to back-substitute, right?
But most textbooks have the answer simplified beyond all recognition.
I'd like to demonstrate how that is done.
We have: .
Factor: .
Back-substitute: .
. . which simplifies to: .
The obvious substitution is u = e^x. Now see here: integrate 1/(x Sqrt[x + 1]) - Wolfram|Alpha(Click on Show steps).