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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Evaluate the following integrals by the method of integration by substitution. Starred exercises may take more problem solving and manipulation than the others. Double starred problems should only be attempted by those who are quite experienced with the Calculus, and may cause anger and frustration. Triple starred problems are reserved for the truly insanely skilled. The beauty of the solution of triple starred problems combined with the sense of accomplishment is a true reward for the hard work put into the problem.

*Last edited by Zhylliolom (2006-08-28 10:32:46)*

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

Good work Ricky, but as you must know, no credit is given for stating only the answer and not the solution method.

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Ugh, you're going to make me type all that up? Fine...

Since 3^{-4x^2} is symmetric across the y-axis.

Let:

Then:

Changing to polar coordinates:

Standard integration follows, we get:

Since 1/81 < 1, 1/81^a^2 approaches 0 as a approaches infinity.

So:

So the answer is:

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

Wait a minute, I never had to use integration by parts...

"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."

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**Zhylliolom****Real Member**- Registered: 2005-09-05
- Posts: 412

substitution*

Monumental effort, Ricky, and your solution is indeed correct. However, you are right in noticing that you didn't ever use any substitution. The solution takes much less work with the proper cleverness of substitution and manipulation and the right knowledge.

*Last edited by Zhylliolom (2006-08-29 09:15:45)*

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**glenn101****Member**- Registered: 2008-04-02
- Posts: 108

1. Let u = 1+x^2

du= 2x dx

= ∫√u du

= u^3/2

------

2

= 2(1+x^2)^3/2

---------------- + c

3

9. = ∫sinx/cosx dx

let u = cosx

du = -sinx dx

=-∫1/u du

= -loge(|cosx|) + c

*Last edited by glenn101 (2009-07-03 16:44:00)*

"If your going through hell, keep going."

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

Hi!

Lets try 17.

where the substitutions used have been z = - ln(x) first, and t^2 = z next.

Jose

*Last edited by juriguen (2009-07-23 23:35:08)*

Make everything as simple as possible, but not simpler. -- Albert Einstein

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

I found 16 is indeed much easier this way:

where for the first step I have used x = exp(ln(x)), and the substitution is then 4x^2 ln(3) = t^2. Finally, the last step is done evaluating the erf function.

Jose

Make everything as simple as possible, but not simpler. -- Albert Einstein

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

Now 18:

using 2 - x = z^2 !

Make everything as simple as possible, but not simpler. -- Albert Einstein

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,664

For #15

Start with the first one and say:

Now for the second integral:

Say:

So

*Last edited by bobbym (2009-07-24 04:06:21)*

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

I would like to propose another integral, which took me really long to solve! (I would grade it at least with **)

If anyone feels this should be another post, since both integration by parts and substitution need to be used, just let me know

Otherwise, have fun with the problem. Here we go:

where

Enjoy!

Make everything as simple as possible, but not simpler. -- Albert Einstein

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

Zhylliolom

I am trying to figure out the last integral, but still struggling to understand a few things:

When you define J(x) you use p, which is any prime? and 1/n... But with the hint the summation changes and has ln(p)...

Also, in the integral to solve, I see dJ(x)... Does this mean that first we should find the differential of J(x)?

Thanks!

Make everything as simple as possible, but not simpler. -- Albert Einstein

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,664

Three easy ones;

*Last edited by bobbym (2009-07-26 03:38:25)*

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**juriguen****Member**- Registered: 2009-07-05
- Posts: 59

This is the only way I see for 19, but it is a little weird!

First consider:

where the substitution used has been alpha x^2 = u^2

Then,

Jose

*Last edited by juriguen (2009-07-25 23:08:29)*

Make everything as simple as possible, but not simpler. -- Albert Einstein

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,664

Hi;

Here is an interesting integral done by substitution.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

And another one:

*Last edited by Identity (2009-08-12 23:11:56)*

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**rzaidan****Member**- Registered: 2009-08-13
- Posts: 59

Hi Identity

sec x + tan x

∫ (1/cos x) dx=∫ sec x dx=∫ sec x ______________ dx

sec x + tan x

(sec x)^2 + sec x tan x

∫ ___________________ dx

sec x + tan x

= ln(sec x + tan x) + C since the numerator is the derivative of the denominator.

Best Regards

Riad Zaidan

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Nice solution rzaidan, but I think the step of multiplying by (secx + tanx) requires a big leap of faith whereas u-substitution does not.

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**rzaidan****Member**- Registered: 2009-08-13
- Posts: 59

Hi Identity

This can be noticed from the derivative of ln(sec x + tan x) + C which is

sec x tan x + sec^2 x sec x ( tan x + sec x )

(d/dx) (ln(sec x + tan x) + C)=__________________ = ___________________ = sec x

secx + tan x sec x + tan x

Best Regards

Riad Zaidan

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**Denominator****Member**- Registered: 2009-11-23
- Posts: 155

Hi. This is my first time posting here.

I'm a year 7 and i think i know the answer to question number 1 and 8!

8 is 1/5 (e^5x)

1 is 2/3 (1+x^2)^(3/2)

May I ask how did you actually get the real forat of the equations? Did you copy from Microsoft Word Equations?

And I also figured out number 6!!!

Is it (1/3)sqrt(2x+1) ?

Please tell me if I'm right

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,664

Hi Denominator;

Welcome to the forum!

I am sorry but that is not correct for number 6:

See my post # 14 above and click #6 for my answer.

*Last edited by bobbym (2009-11-23 19:12:33)*

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**Denominator****Member**- Registered: 2009-11-23
- Posts: 155

Okay my bad.

How do you make the actual equation insetad of just typing it up in text?

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**Identity****Member**- Registered: 2007-04-18
- Posts: 934

Ooops sorry rzaidan I was going to post my solution but I forgot

*Last edited by Identity (2009-11-24 10:50:48)*

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**123ronnie321****Member**- Registered: 2010-09-28
- Posts: 128

bobbym wrote:

Hi;

Here is an interesting integral done by substitution.

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