Effortlessly Integrate Complex Functions: Learn How Here!

[paulmdrdo1]

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]

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?

 

[paulmdrdo1]

i get
$\displaystyle -2\int\frac{v}{u}dv$ --- what's next?

 

[MarkFL]

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?

 

[paulmdrdo1]

this is what i get

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

 

[MarkFL]

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?

 

[paulmdrdo1]

$\displaystyle u=(v-1)^3+1$ - how did you get this?

 

[MarkFL]

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

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

 

[paulmdrdo1]

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]

Yes, this is what I have as well. Now what do we normally do with a quadratic in the denominator?

 

[paulmdrdo1]

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]

Excellent! (Sun)

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

 

[paulmdrdo1]

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]

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?

 

[paulmdrdo1]

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]

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.

 

[paulmdrdo1]

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]

You are correct, but you need to undo one last substitution, that is, to replace $v$ using the substitution we initially used.

 

[paulmdrdo1]

How about this
$\displaystyle -\frac{4}{\sqrt{3}}tan^{-1}\frac{2(u-1)^{\frac{1}{3}}-1}{\sqrt{3}}+C$ -- correction

 

[MarkFL]

What happened to the factor of and the inverse tangent function?

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

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

 

[paulmdrdo1]

yes i forgot to write the factor which i edited. :)(Clapping)thanks markFl i finally made it!

 

[MarkFL]

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]

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)

 

[paulmdrdo1]

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]

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...