Integration

yonski · 2007-08-10 05:00:00

Hi there,
i'm getting frustrated by a lot of the integration material in my latest maths book. I feel like I understand all the topics but there are some questions which just throw me completely. I'll just list two here:

1. Show that

2. Given that

show that, for n ≥ 2,

The second question obviously involves integration by reduction, but I can't for the life of me see how to work either of them out.

Any help appreciated!

mikau · 2007-08-10 07:20:29

for problem 1. let u = sqrt(1 + x), then u^2 = x + 1, so x + 5 must be u^2 + 4, we also have dx = 2u du

so the problem becomes

∫ (2u^2) / (u^2 + 4) du evaluated from x = 3 to 8, which means, u = 2 to 3

we have that 2u^2 = 2u^2 + 8 - 8 which is 2(u^2 + 4) - 8 so the integral becomes

∫ (2(u^2 + 4) - 8 )/ (u^2 + 4) from u = 3 to 8

= ∫ 2 - ∫ 8/(u^2 + 4) du, now its a fairly easy trig substitution problem.

for the second, we are not actually solving the integral, just expressing it in terms of another (in this case, slightly reduced) integral. This just screams INTEGRATION BY PARTS!

so lets try. Let u = (1 + x^2)^-n and dV = dx, then du = -2nx/ (1 + x^2)^(n +1), and V = x so the integral becomes

x(1 + x^2)^-n + ∫ 2nx^2/(1 + x^2)^(n+1) now i'm going to use the same trick I did in the problem above, 2nx^2 = 2nx^2 + 2n - 2n, so we have ∫ 2n(x^2 + 1)/(x^2 n + 1)^n+1 - 2n ∫1/(x^2 n + 1)^(n+1) which translates to ∫ 2n/(x^2 +1)^n - 2n ∫1/(x^2 n + 1)^(n+1).

Ugh.. PUTTING IT ALL TOGETHER

I(n) = x(1 + x^2)^-n + ∫ 2n/(x^2 +1)^n - 2n ∫1/(x^2 n + 1)^(n+1)

note the first portion, before the integrals, can be evaluated from 0 to 1 which gives us 2^-n, now if we look closely, we find that the equation becomes

I(n) = 2^-n + 2n I(n) - 2n I(n+1)

rearranging we get 2n I (n+1) = 2^-n + (2n - 1)I(n)

now note we have I(n+1) and I(n) instead of n and n-1, but n is just an arbitrary number so lets just reindex. (in otherwords, replace n+1 with n and n with n-1 then we get

2(n-1)I(n) = 2^-(n-1) + (2n - 2 - 1)I(n-1)

so

2(n-1) I(n) = 2^(1 - n) + (2n - 3)I(n-1) QED.

phew.. I need a cold soda..

Last edited by mikau (2007-08-10 07:22:54)

mikau · 2007-08-10 07:30:40

you should note that whenever you have an integral of the form:

∫ka/(a + b)^n you can always write this as ∫(ka + kb - kb)/(a + b)^n = ∫ k/(a + b)^(n-1) - ∫kb/(a + b)^n it doesn't hurt it since you are just adding 0 to the numerator and it can get you out of some tight spots. It tends to come up in trig substitution problems where you have something like

∫ 2x^2/(1 + x^2) dx just add + 2 and -2 to the numerator and you can separate it as ∫1dx + ∫2/(1 + x^2)dx = x + 2 arctan(x) + C. wasn't that simple?

yonski · 2007-08-10 08:27:30

Ah okay, cool. Just ran through that on paper and I get those now, thanks! And learnt a few little tricks there which i'll try to remember.

Hopefully i'll be able to have a better crack at some of the others now. I think the biggest problem I have is not knowing whether to try integration by parts or by substitution, and if so, what substitution to use. Hopefully it'll become easier the more questions I do.

Once again, thanks.

mikau · 2007-08-10 09:10:55

well obviously in some cases you can use either, and in some cases you must use both!! x_x

what I reccomend is, always consider u substitution first, the best way is to look for the derivative of some inner function as a factor, then see if substitution works, if you don't see one, try another method. If you do see the derivative of an inner function as a factor, but substitution gets you nowhere, in otherwords, you end up with an integral of the form f(x)f'(x)g(x) dx then integration by parts is obviously worth a shot.

When the square root of a binomial comes in, usually its a trig substitution.

When its something of the form 4/[(x+5)(x+3)] dx, it usually involves partial fractions.

Just a few tips.

Krizalid · 2007-08-13 12:11:55

mikau wrote:

When its something of the form 4/[(x+5)(x+3)] dx, it usually involves partial fractions.

Well sometimes it can be avoided.

akshay1124 · 2012-03-19 04:49:27

some imp tips while solving integration?

bobbym · 2012-03-19 08:39:37

Hi akshay1124;

Welcome to the forum!

1)Practice! Practice! Practice!

2)Substitute!

3)IBP!

4)DUIS!

5)EMS!

5)Tables and Mathematica or Wolfram!

zetafunc. · 2012-03-19 09:10:37

bobbym wrote:

Hi akshay1124;
Welcome to the forum!
1)Practice! Practice! Practice!
2)Substitute!
3)IBP!
4)DUIS!
5)EMS!
5)Tables and Mathematica or Wolfram!

What are 4 and 5? Sorry, I haven't heard those terms before... or maybe I am too stupid to work out what they stand for, haha.

bobbym · 2012-03-19 09:15:32

Hi;

Five is the Euler Mclaurin summation formula. Tables are giant books of integrals. Mathematica is a Computer Algebra System. Wolfram Alpha is Mathematica online for everyone to use.

anonimnystefy · 2012-03-19 09:18:59

But what is duis?

bobbym · 2012-03-19 09:20:37

DUIS - Differentiation under the integral sign. You can read Richard Feynman's comments about it.

Math Is Fun Forum

#1 2007-08-10 05:00:00

Integration

#2 2007-08-10 07:20:29

Re: Integration

#3 2007-08-10 07:30:40

Re: Integration

#4 2007-08-10 08:27:30

Re: Integration

#5 2007-08-10 09:10:55

Re: Integration

#6 2007-08-13 12:11:55

Re: Integration

#7 2012-03-19 04:49:27

Re: Integration

#8 2012-03-19 08:39:37

Re: Integration

#9 2012-03-19 09:10:37

Re: Integration

#10 2012-03-19 09:15:32

Re: Integration

#11 2012-03-19 09:18:59

Re: Integration

#12 2012-03-19 09:20:37

Re: Integration

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