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#### paulmdrdo

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- May 13, 2013

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

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- May 13, 2013

- 386

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

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

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

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- May 13, 2013

- 386

i get

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

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

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- #4

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?i get

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

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- #5

- May 13, 2013

- 386

this is what i get

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

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

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- #6

You cannot simply replace $u$ with $v$ because you did not use that as your substitution. Let's go back to:this is what i get

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

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

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?

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- May 13, 2013

- 386

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

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- #8

I took our substitution:$\displaystyle u=(v-1)^3+1$ - how did you get this?

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

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:

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

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- #9

- May 13, 2013

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

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- #11

- May 13, 2013

- 386

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

what's next?

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- #12

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- #13

- May 13, 2013

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we can use trig substitution butExcellent!

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

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?

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- #15

- May 13, 2013

- 386

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

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- #16

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

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

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

and now you have:

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

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

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- #17

- May 13, 2013

- 386

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

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- #18

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- #19

- 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

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

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- #20

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

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

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- #21

- May 13, 2013

- 386

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

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- #22

Good job! And congratulations on sticking with it to the end.yes i forgot to write the factor which i edited. thanks markFl i finally made it!

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- #23

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

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- May 13, 2013

- 386

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?

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

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- #25

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

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...