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# FP2 Integration watch

1. Could someone explain how to do the following question please?

It actually has 2 parts before this one:

and

which are both fairly easy to do, but I can't put them together to get the third one.

Any help is much appreciated.
2. i imagine its some 'by parts' trickery. but i cant see what it is. sorry
3. See the attached file.

Note however that I got the integral circled in red from the Integrator and couldn't do it manually. I don't think we can do it based on FP2 integration. If I'm wrong let me know how to do it!
Attached Images

4. Yeah. I could get that integral but got stuck there too.

Hmmm... the question is from the Heinemann FP2 textbook.

I wonder if a trig substitution is needed...
5. How to get the thing circled in red, or how to actually integrate it?
6. How to actually integrate it
7. I think I can do it, but I don't know how to get the mathematical symbols when I reply.

You have to use a hyperbolic substitution of x = sinh u (I think) I haven't actually done it yet.
8. Just integrate by parts and it comes out immediately.
9. Yeah that works. Thanks.

EDIT: I was talking to Handy
10. I think using x = sinh u will give the answer.

x = sinh u
so the function in the square root becomes (cosh u)^2

So basically find the integral of cosh u and then substitute: x = sinh u

Is this correct?
11. Okay, so it correct!
12. Well once you've substituted x = sinh u and simplified, the integral becomes cosh^2 u.

Using an identity you can finish it off.
13. (Original post by Wrangler)
Just integrate by parts and it comes out immedeately.
Yeah, the problem was I couldn't integrate √(x^2 + 1) but I've figured out how to do it now.
14. Ah, ok - sorry about that. I think the most direct approach would be to use:

x²/√(x²+1) = √(x²+1) - 1/√(x²+1)

So, integrate by parts to get:

I = ∫x[x/√(x²+1)]dx = x√(x²+1) - ∫√(x²+1)dx = x√(x²+1) - ∫x[x/√(x²+1)]dx - ∫[1/√(x²+1)]dx = x√(x²+1) - arcsinh(x) - I

And we're done.
15. (Original post by Wrangler)
Ah, ok - sorry about that. I think the most direct approach would be to use:

x²/√(x²+1) = √(x²+1) - 1/√(x²+1)

So, integrate by parts to get:

I = ∫x[x/√(x²+1)]dx = x√(x²+1) - ∫√(x²+1)dx = x√(x²+1) - ∫x[x/√(x²+1)]dx - ∫[1/√(x²+1)]dx = x√(x²+1) - arcsinh(x) - I

And we're done.
That weird 'I' method isn't on the syllabus.
It's easier to use a substitution, and he already has.
16. (Original post by Handy)
That weird 'I' method isn't on the syllabus.
It's easier to use a substitution, and he already has.
There's nothing weird about the "I", I've just used it to save me writing out the integral several times. I personally think it's a little "slicker" to avoid using extra substitutions. A quick rearrangement of my line of working gives:

I = ½[x√(x²+1) - arcsinh(x)]

17. The "I" method is in the (Edexcel) syllabus by my reckoning. Especially considering that Reduction Formulae is the next exercise after the one I'm doing.
18. I only stands for Integral ie, some function=I. It'll come in handy when you need to move the integral and the integrand from side to side, in recurrence intergration.
19. Right, New question:

20. (Original post by e-unit)
The "I" method is in the (Edexcel) syllabus by my reckoning. Especially considering that Reduction Formulae is the next exercise after the one I'm doing.
Fair enough.

I don't need to know it

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