I can see that

is easier to integrate, and can carry that integration out.

However, I can't see how your:

corresponds to

or how you got there.

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- Mar 14th 2011, 02:55 PMMaverickUK82
I can see that

is easier to integrate, and can carry that integration out.

However, I can't see how your:

corresponds to

or how you got there. - Mar 14th 2011, 04:40 PMtopsquark
This is as step by step as I can make it.

The original integral is

It's always a good idea to try to get the denominator to be as simple as possible. (Though there are exceptions to that.) So this gives the idea of the substitution

To sub this back into the integral we need to know what v is. So solving for v in terms of u we get:

But we are not done. We also need to find an expression for dv in terms of u also. So

Solving this for dv gives

Now we need to sub in our values for v and dv into the original integral:

Factoring out the constants gives

And subbing back in gives

-Dan - Mar 15th 2011, 12:24 PMMaverickUK82
THanks topsquark... Whilst I am able to follow that through, and can see what is being done. I would never have came up with that on my own. THanks for your patience and help!

- Mar 15th 2011, 06:24 PMtopsquark
- Mar 22nd 2011, 03:15 PMMaverickUK82
- Mar 22nd 2011, 03:46 PMtopsquark
- Mar 23rd 2011, 08:28 AMMaverickUK82
Topsquark, thanks again, I have, hopefully, one final question with regards to the above.

We have

yet, we have

I'm struggling with this, why does the derivative of v equal one third of the derivative of u (which of course would be 1)?

I think with a bit of clarification on this I will be done! I really appreciate your patience. - Mar 23rd 2011, 04:50 PMtopsquark
- Mar 24th 2011, 01:39 AMMaverickUK82
- Mar 25th 2011, 03:52 PMMaverickUK82
Topsquark.

Thanks for your help on this. I am still confused with the last bit (last post). I have found some good material on integration with by substitution and will continue to read up on it and hopefully it will become clearer. - Mar 25th 2011, 08:56 PMtopsquark