Homework Help: Integration by parts tanx help

carbz · May 1, 2008

I actually have two here, so I will just list both:

1. The problem statement, all variables and given/known data
[tex]\int\frac{x}{x^{2}+4x+4}dx[/tex]

2. Relevant equations
None

3. The attempt at a solution
I tried this one twice. I honestly have no idea how to do it, and I used integration by parts. The first time, I reduced it down to:
[tex]\int\frac{1}{x} + \frac{1}{4} + \frac{x}{4}dx[/tex]
But, this is wrong.

I tried it a second time by using integration by parts to obtain:
[tex]\int\frac{x}{(x+2)(x+2)}dx[/tex], then I reduced that down, since integration by parts does not work. So, I was hoping to know what I am susposed to do.

The second one is a bit different:
1. The problem statement, all variables and given/known data
[tex]/int[/tex] [tex](tan^{2}(x))dx[/tex]

2. Relevant equations
[tex]tan^{2}(x) + 1 = sec^{2}[/tex]

3. The attempt at a solution
I used the regular formula that I listed to get:
[tex]\int(sec^{2}(x) - 1)dx[/tex].
I just integrated to: [tex] tan^{2} - x + c [/tex]
I wanted to see if this one is correct.

Thankyou for your help.

drdizzard · May 1, 2008

For the first problem you need to use integration by partial fractions to solve it.

You were on the right track in the second problem changing (tanx)^2 to (secx)^2 - 1 now you just have to integrate each seperatly which you did right except for the integral of (secx)^2 is just tanx

carbz · May 1, 2008

I did use that too for the first one, and I could not find an answer. When I did it, I got 2=0 which is not an answer.

All right, thank you for the second one.

jambaugh · May 1, 2008

Quicker solution to the first problem:

Use variable substitution: [tex] u = x+2[/tex]
You'll then get:
[tex] \int \frac{something}{u^2}du[/tex]
which you can then separate into powers of u and integrate.

Your integration in the second problem is not quite correct.
Recall that [tex]\frac{d}{dx} \tan (x) = \sec^2(x)[/tex] not tan squared!!!

flebbyman · May 1, 2008

You got [tex]
\int\frac{x}{(x+2)^2}dx[/tex]

use the substitution u=x+2, so x=u-2 and dx=du and take it from there

carbz · May 1, 2008

What both of you are saying will produce:
[tex]\int\frac{x}{u^{2}}[/tex]
Shouldn't the x in the numerator be apart of it too?

jambaugh · May 1, 2008

carbz said: ↑

What both of you are saying will produce:
[tex]\int\frac{x}{u^{2}}[/tex]
Shouldn't the x in the numerator be apart of it too?

You must finish the substitution.... rewrite x in terms of u.

[edit: And of course when you're done integrating, rewrite your answer back in terms of x.]

carbz · May 1, 2008

yes, you said that [tex]u = x+2[/tex]
The derivative of that is [tex]du = 1dx[/tex]

That would only get:
[tex]\int\frac{x}{u^{2}}[/tex]

Then you would solve x equal to u-2?

That would get:

[tex]\int\frac{u-2}{u^{2}}[/tex]

That would make:
[tex]\int\frac{1}{u} - \int\frac{2}{u^{2}}[/tex] = [tex]ln(x) - \frac{-2}{u}[/tex]
Then, of course, you would put x+2 back in for all u's. Right?

jambaugh · May 1, 2008

Right! (But don't forget the +C!!!) ;-)

tiny-tim · May 1, 2008

Hi carbz!

A partial fractions method (which avoids having to substitue and then substitute back again ) is:

[tex]\int\frac{x}{x^{2}+4x+4}dx[/tex]

[tex]=\,\int\frac{x+2}{x^{2}+4x+4}dx
\,-\,\int\frac{2}{x^{2}+4x+4}dx[/tex]

jambaugh · May 1, 2008

tiny-tim said: ↑

Hi carbz!

A partial fractions method (which avoids having to substitue and then substitute back again ) is:

[tex]\int\frac{x}{x^{2}+4x+4}dx[/tex]

[tex]=\,\int\frac{x+2}{x^{2}+4x+4}dx
\,-\,\int\frac{2}{x^{2}+4x+4}dx[/tex]

You've not solved it yet. You'll still need to do a substitution to integrate the partial fractions. Now you must do it twice. Also since its a linear substitution you can do it in your head once you're practiced.

rootX · May 1, 2008

jambaugh said: ↑

You've not solved it yet. You'll still need to do a substitution to integrate the partial fractions. Now you must do it twice. Also since its a linear substitution you can do it in your head once you're practiced.

his thing = (x+2)/(x+2)^2 - 2/(x+2)^2
so, no *twice* substitution

tiny-tim · May 1, 2008

rootX said: ↑

… so, no *twice* substitution

No … jambaugh is right … you and I can do it in our heads, but without that knack it does require a substitution!

tron_2.0 · May 1, 2008

actually couldnt you do it with partial fractions and integrating once?

you can break up x/(x^2+4x+4) into x/(x+2)^2

therefore by partial fractions you could break it into [x/(x+2)^2]= A/(x+2) + Bx+C/[(x+2)^2]

from there you can solve for A, B, and C and plug that back into your original integral

not sure if thats what you originally ment but idk.

tiny-tim · May 1, 2008

tron_2.0 said: ↑

… therefore by partial fractions you could break it into [x/(x+2)^2]= A/(x+2) + Bx+C/[(x+2)^2] …

Hi tron_2.0!

Yes … and B will be 0 …

… but that's exactly what we have done!

tron_2.0 · May 1, 2008

ahh my mistake, i should have read the replies more closely =[

flebbyman · May 1, 2008

Nice one carbz

jambaugh · May 1, 2008

OK I take back the "twice", as it is the same substitution and you can write the two summed integrals as a single one.

In effect it is the same two steps in different orders as when I say "you can then separate into powers of u and integrate" this is a trivial case of expansion into partial fractions. Ultimately the preference is one of which you can do more easily the first substitution or the first expansion by partial fractions.

carbz · May 1, 2008

well, the C is always annoying, but thank you.