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Integration problem.

paulmdrdo

Active member
May 13, 2013
386
any idea how to integrate this one

$\displaystyle \int(\frac{-2}{3u(u-1)^{\frac{1}{3}}}-\frac{2}{3u(u-1)^{\frac{2}{3}}})du$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I think I would begin by factoring a -2 out front and combining terms in the integrand to get:

\(\displaystyle -2\int\frac{(u-1)^{\frac{1}{3}}+1}{3u(u-1)^{\frac{2}{3}}}\,du\)

Now, try letting the numerator be another variable, such as $v$...what do you find?
 

paulmdrdo

Active member
May 13, 2013
386
i get
$\displaystyle -2\int\frac{v}{u}dv$ --- what's next?
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
i get
$\displaystyle -2\int\frac{v}{u}$ --- what's next?
You are on the right track, but you need to express $u$ in terms of $v$ and you need a differential. Now what do you have?
 

paulmdrdo

Active member
May 13, 2013
386
this is what i get

$\displaystyle -2\int\frac{v}{v}dv \,=\,-2v\,=\,-2(u-1)^{\frac{1}{3}}$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
this is what i get

$\displaystyle -2\int\frac{v}{v}dv \,=\,-2v\,=\,-2(u-1)^{\frac{1}{3}}$
You cannot simply replace $u$ with $v$ because you did not use that as your substitution. Let's go back to:

\(\displaystyle -2\int\frac{(u-1)^{\frac{1}{3}}+1}{3u(u-1)^{\frac{2}{3}}}\,du\)

Now, I am assuming you let:

\(\displaystyle v=(u-1)^{\frac{1}{3}}+1\,\therefore\,dv=\frac{1}{3}(u-1)^{-\frac{2}{3}}\,du\)

Now, this takes care of everything except that $u$ in the denominator, and so using our substitution, we solve for $u$ to obtain:

\(\displaystyle u=(v-1)^3+1\)

and so now we have:

\(\displaystyle -2\int\frac{v}{(v-1)^3+1}\,dv\)

Now, factor the denominator as the sum of two cubes...you should find that a cancellation becomes possible...what do you have?
 

paulmdrdo

Active member
May 13, 2013
386
$\displaystyle u=(v-1)^3+1$ - how did you get this?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
$\displaystyle u=(v-1)^3+1$ - how did you get this?
I took our substitution:

\(\displaystyle v=(u-1)^{\frac{1}{3}}+1\)

and solved it for $u$. First we subtract through by $1$ to get:

\(\displaystyle v-1=(u-1)^{\frac{1}{3}}\)

Cube both sides:

\(\displaystyle (v-1)^3=u-1\)

Add $1$ to both sides:

\(\displaystyle (v-1)^3+1=u\)
 

paulmdrdo

Active member
May 13, 2013
386
following your instruction this is what i have

$\displaystyle -2\int\frac{v}{v(v^2-3v+3)}dv\,=\,-2\int\frac{dv}{(v^2-3v+3)}$ is my factorization right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, this is what I have as well. Now what do we normally do with a quadratic in the denominator?
 

paulmdrdo

Active member
May 13, 2013
386
completing the square

i would have this $\displaystyle -2\int\frac{dv}{(v-\frac{3}{2})^2+(\sqrt{\frac{3}{4}})^2}$
what's next?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Excellent! (Sun)

Now, when we have the sum of two squares in the denominator, what type of substitution do we normally use?
 

paulmdrdo

Active member
May 13, 2013
386
Excellent! (Sun)

Now, when we have the sum of two squares in the denominator, what type of substitution do we normally use?
we can use trig substitution but
it is also leading to an inverse tangent.

$\displaystyle -2\int\frac{dv}{(v-\frac{3}{2})^2+(\sqrt{\frac{3}{4}})^2}\,=\,\frac{1}{(v-\frac{3}{2})}\,tan^{-1}\frac{\frac{\sqrt{3}}{2}}{(v-\frac{3}{2})}\,=\,\frac{2}{2v-3}tan^{-1}\frac{\sqrt{3}}{2v-3}$ is this correct?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
It can lead to the inverse tangent function because a good substitution in this case involves the tangent function. We want to let:

\(\displaystyle v-\frac{3}{2}=\frac{\sqrt{3}}{2}\tan(\theta)\)

Do you see why?
 

paulmdrdo

Active member
May 13, 2013
386
It can lead to the inverse tangent function because a good substitution in this case involves the tangent function. We want to let:

\(\displaystyle v-\frac{3}{2}=\frac{\sqrt{3}}{2}\tan(\theta)\)

Do you see why?
what i have in mind is this formula $\displaystyle \int \frac{du}{a^2+u^2}\,=\,\frac{1}{a}tan^{-1}\frac{u}{a} +C$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
what i have in mind is this formula $\displaystyle \int \frac{du}{a^2+u^2}\,=\,\frac{1}{a}tan^{-1}\frac{u}{a} +C$
Okay, if you are going to use that formula, then you want to let:

\(\displaystyle w=v-\frac{3}{2}\,\therefore\,dw=dv\)

\(\displaystyle a=\frac{\sqrt{3}}{2}\)

and now you have:

\(\displaystyle -2\int\frac{dw}{w^2+a^2}\)

So, apply your formula, then undo all the substitutions.
 

paulmdrdo

Active member
May 13, 2013
386
this is my answer

$\displaystyle -2\int\frac{dw}{w^2+a^2}=-2\,\frac{1}{\frac{\sqrt{3}}{2}}tan^{-1}\frac{\frac{2v-3}{2}}{\frac{\sqrt{3}}{2}}\,=\,-\frac{4}{\sqrt{3}}tan^{-1}\frac{2v-3}{\sqrt{3}}+C$
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
You are correct, but you need to undo one last substitution, that is, to replace $v$ using the substitution we initially used.
 

paulmdrdo

Active member
May 13, 2013
386
How about this
$\displaystyle -\frac{4}{\sqrt{3}}tan^{-1}\frac{2(u-1)^{\frac{1}{3}}-1}{\sqrt{3}}+C$ -- correction
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
What happened to the factor of and the inverse tangent function?

You want, as you final statement, to write something like:

\(\displaystyle \int\left(\frac{-2}{3u(u-1)^{\frac{1}{3}}}-\frac{2}{3u(u-1)^{\frac{2}{3}}} \right)du=-\frac{4}{\sqrt{3}}\tan^{-1}\left(\frac{2\sqrt[3]{u-1}-1}{\sqrt{3}} \right)+C\)
 

paulmdrdo

Active member
May 13, 2013
386
yes i forgot to write the factor which i edited. :)(Clapping)thanks markFl i finally made it!
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
yes i forgot to write the factor which i edited. :)(Clapping)thanks markFl i finally made it!
Good job! And congratulations on sticking with it to the end. (Clapping)
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Okay, as a follow-up...can you derive the formula you used?

\(\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a} \right)+C\) where $a$ is a real non-zero constant.

My philosophy has always been, if you can derive it, then you own it, and can use it. (Nerd)
 

paulmdrdo

Active member
May 13, 2013
386
Okay, as a follow-up...can you derive the formula you used?

\(\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a} \right)+C\) where $a$ is a real non-zero constant.

My philosophy has always been, if you can derive it, then you own it, and can use it. (Nerd)
(Envy) that would be interesting. my teacher in calculus didn't teach us how derive that formula. but my approach to that would be using trig-substitution. am i right?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Yes, you are correct, we can use a trigonometric substitution...

\(\displaystyle u=a\tan(\theta)\,\therefore\,du=a\sec^2(\theta)\,d\theta\)

and we have:

\(\displaystyle \int\frac{a\sec^2(\theta)}{a^2\tan^2(\theta)+a^2} \,d\theta\)

\(\displaystyle \frac{1}{a}\int\frac{\sec^2(\theta)}{\tan^2(\theta)+1} \,d\theta\)

Now, by Pythagoras, we know:

\(\displaystyle \tan^2(\theta)+1=\sec^2(\theta)\)

and now we have:

\(\displaystyle \frac{1}{a}\int\,d\theta=\frac{1}{a}\theta+C\)

Next, we want to write $\theta$ in terms of $u$, and so we go back to:

\(\displaystyle u=a\tan(\theta)\,\therefore\,\theta=\tan^{-1}\left(\frac{u}{a} \right)\)

and so we may state:

\(\displaystyle \int\frac{du}{u^2+a^2}=\frac{1}{a}\tan^{-1}\left(\frac{u}{a} \right)+C\)

See if you can derive an equivalent formula using the cotangent function...