Integration by Substitution

Zhylliolom · 2006-08-27 14:00:29

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)

Ricky · 2006-08-27 14:50:07

Zhylliolom · 2006-08-27 15:27:02

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

Ricky · 2006-08-27 15:45:29

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:

Ricky · 2006-08-27 16:02:39

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

Zhylliolom · 2006-08-27 16:32:21

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)

glenn101 · 2009-07-03 16:42:13

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)

juriguen · 2009-07-23 23:34:26

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)

juriguen · 2009-07-23 23:50:05

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

juriguen · 2009-07-24 01:44:07

Now 18:

using 2 - x = z^2 !

bobbym · 2009-07-24 03:45:00

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)

juriguen · 2009-07-24 04:01:11

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!

juriguen · 2009-07-24 05:51:57

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!

bobbym · 2009-07-25 04:33:33

Three easy ones;

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

juriguen · 2009-07-25 19:46:35

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)

bobbym · 2009-07-29 12:14:36

Hi;

Here is an interesting integral done by substitution.

Identity · 2009-08-12 23:09:02

And another one:

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

rzaidan · 2009-08-13 23:08:34

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

Identity · 2009-08-14 02:57:44

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

rzaidan · 2009-08-15 02:12:30

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

Denominator · 2009-11-23 19:03:37

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

bobbym · 2009-11-23 19:10:55

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)

Denominator · 2009-11-24 05:53:34

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

Identity · 2009-11-24 10:50:15

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

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

123ronnie321 · 2010-09-28 07:13:18

bobbym wrote:

Hi;
Here is an interesting integral done by substitution.

Ans - 0

Math Is Fun Forum

#1 2006-08-27 14:00:29

Integration by Substitution

#2 2006-08-27 14:50:07

Re: Integration by Substitution

#3 2006-08-27 15:27:02

Re: Integration by Substitution

#4 2006-08-27 15:45:29

Re: Integration by Substitution

#5 2006-08-27 16:02:39

Re: Integration by Substitution

#6 2006-08-27 16:32:21

Re: Integration by Substitution

#7 2009-07-03 16:42:13

Re: Integration by Substitution

#8 2009-07-23 23:34:26

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#10 2009-07-24 01:44:07

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#11 2009-07-24 03:45:00

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#13 2009-07-24 05:51:57

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#14 2009-07-25 04:33:33

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#15 2009-07-25 19:46:35

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#16 2009-07-29 12:14:36

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#18 2009-08-13 23:08:34

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#19 2009-08-14 02:57:44

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#20 2009-08-15 02:12:30

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#21 2009-11-23 19:03:37

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#22 2009-11-23 19:10:55

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#23 2009-11-24 05:53:34

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#24 2009-11-24 10:50:15

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#25 2010-09-28 07:13:18

Re: Integration by Substitution

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