Integration (2 Viewers)

pikachu975 · Feb 14, 2017

How do you integrate xdx/sqrt(x-x^2) (exercise 4.3 question 1h) from terry lee)?

Thanks for the help

Drongoski · Feb 14, 2017

pikachu975 · Feb 14, 2017

Drongoski said:
$\int \frac {x}{\sqrt{x-x^2}}dx = \int \frac {(-\frac {1}{2}(1-2x) + \frac {1}{2})}{\sqrt{x-x^2}} dx = -\frac{1}{2}\int \frac {1-2x}{\sqrt {x-x^2}}dx + \frac {1}{2} \int \frac {1}{\sqrt {\frac {1}{4} - (x^2-x+ \frac {1}{4})}} dx\\ \\ = -\frac{1}{2}\int (x-x^2)^{-\frac{1}{2}}d(x-x^2) + \frac{1}{2}\int\frac {1}{\sqrt {(\frac{1}{2})^2 - (x-\frac{1}{2})^2}} dx\\ \\ = -\sqrt{x-x^2} + \frac {1}{2} sin^{-1}\frac {x-\frac{1}{2}}{\frac {1}{2}} + C\\ \\ = -\sqrt{x-x^2} + \frac{1}{2} sin^{-1}(2x-1)+ C$

Thanks for the help. Do you mind explaining your thought process for getting the first line where you knew to do -0.5 (1-2x) + 0.5? Thanks again

BenHowe · Feb 14, 2017

Hey I know it's been answered but I still wanna upload it so I can ask a question.

. How can I put the latex in the actual post like the others above? I only know how to link as a picture

pikachu975 · Feb 14, 2017

BenHowe said:
Hey I know it's been answered but I still wanna upload it so I can ask a question. View attachment 33794. How can I put the latex in the actual post like the others above? I only know how to link as a picture

Did you type it as a code? Click reply to Drongoski's post and see the code required for latex

BenHowe · Feb 14, 2017

Ok lemme try $$\int\frac{x}{\sqrt{x-x^{2}}}dx$ $=-\frac{1}{2}\int\frac{-2x}{\sqrt{-x^{2}+x}}dx$ $=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{-x^{2}+x}}dx$ $ now -x^{2}+x=-(x^{2}-x)$ $=-((x-\frac{1}{2})^{2}-\frac{1}{4})$ $=(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}$ $=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}}}dx$ $=-\sqrt{-x^{2}+x}+\frac{1}{2}sin^{-1}(2x-1)+c\hspace{0.1cm}$ [\tex]$

pikachu975 · Feb 14, 2017

BenHowe said:
Ok lemme try $$\int\frac{x}{\sqrt{x-x^{2}}}dx$ $=-\frac{1}{2}\int\frac{-2x}{\sqrt{-x^{2}+x}}dx$ $=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{-x^{2}+x}}dx$ $ now -x^{2}+x=-(x^{2}-x)$ $=-((x-\frac{1}{2})^{2}-\frac{1}{4})$ $=(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}$ $=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}}}dx$ $=-\sqrt{-x^{2}+x}+\frac{1}{2}sin^{-1}(2x-1)+c\hspace{0.1cm}$ [\tex][/QUOTE] You don't press enter for a new line maybe it is \\$

BenHowe · Feb 14, 2017

Drongoski · Feb 14, 2017

pikachu975 said:
Thanks for the help. Do you mind explaining your thought process for getting the first line where you knew to do -0.5 (1-2x) + 0.5? Thanks again

Using my personal way of integrating, I note that the derivative of x-x² is 1-2x. so I prepare the denominator of the integrand to have a multiple of 1-2x with some remainder. What I have done is to fiddle the integrand into 2 parts that are easily integrable. Because I have been doing this for many many many years, I can see what I need to do.

pikachu975 · Feb 14, 2017

Drongoski said:
Using my personal way of integrating, I note that the derivative of x-x² is 1-2x. so I prepare the denominator of the integrand to have a multiple of 1-2x with some remainder. What I have done is to fiddle the integrand into 2 parts that are easily integrable. Because I have been doing this for many many many years, I can see what I need to do.

How would you integrate x^2 dx / (x^2 - 2x + 3) lol sorry integration is so hard....

I got x + ln|x^2 - 2x+3| - (tan^-1 (x-1 / root2))/root 2 but it looks weird

InteGrand · Feb 14, 2017

pikachu975 said:
Thanks for the help. Do you mind explaining your thought process for getting the first line where you knew to do -0.5 (1-2x) + 0.5? Thanks again

$\noindent Steps to evaluating any integral like this (linear function on numerator and degree two polynomial (quadratic) in denominator or in a square root in denominator) is as follows. Let $q(x)$ be the quadratic present (it was $x -x^{2}$ in your case).$

$\noindent 1. Write the linear function in the numerator as $aq'(x) + b$ for some suitable constants $a$ and $b$. (You can show as an exercise that there'll always exist a unique pair of such constants $a$ and $b$.)$

$\noindent 2. Observe the integral becomes $a \int\frac{q'(x)}{\sqrt{q(x)}}\,\mathrm{d}x + b\int\frac{1}{\sqrt{q(x)}}\,\mathrm{d}x$. (Or leave out the square roots obviously if the original integral had no square roots.)$

$\noindent 3. Evaluate these two integrals. The first can be done by inspection or substituting $u = q(x)$, $\mathrm{d}u = q'(x)\,\mathrm{d}x$. The second can be done e.g. by completing the square to make it of the form of a standard integral (though it may not be on the Standard Integrals sheet anymore).$

InteGrand · Feb 14, 2017

pikachu975 said:
How would you integrate x^2 dx / (x^2 - 2x + 3) lol sorry integration is so hard....

I got x + ln|x^2 - 2x+3| - (tan^-1 (x-1 / root2))/root 2 but it looks weird

$\noindent Add and subtract $-2x +3$ to the numerator of the integrand, so the integral becomes $\int \left(1 + \frac{2x-3}{x^{2} -2x +3}\right)\,\mathrm{d}x = x + \int \frac{2x-3}{x^2 -2x +3}\,\mathrm{d}x$. This latter integral can be tackled using the procedure outlined above.$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent Steps to evaluating any integral like this (linear function on numerator and degree two polynomial (quadratic) in denominator or in a square root in denominator) is as follows. Let $q(x)$ be the quadratic present (it was $x -x^{2}$ in your case).$

$\noindent 1. Write the linear function in the numerator as $aq'(x) + b$ for some suitable constants $a$ and $b$. (You can show as an exercise that there'll always exist a unique pair of such constants $a$ and $b$.)$

$\noindent 2. Observe the integral becomes $a \int\frac{q'(x)}{\sqrt{q(x)}}\,\mathrm{d}x + b\int\frac{1}{\sqrt{q(x)}}\,\mathrm{d}x$. (Or leave out the square roots obviously if the original integral had no square roots.)$

$\noindent 3. Evaluate these two integrals. The first can be done by inspection or substituting $u = q(x)$, $\mathrm{d}u = q'(x)\,\mathrm{d}x$. The second can be done e.g. by completing the square to make it of the form of a standard integral (though it may not be on the Standard Integrals sheet anymore).$

Thanks a lot for this. How about with a higher degree on the numerator - e.g. x^3 dx / (x^2 + x + 1)

InteGrand · Feb 14, 2017

pikachu975 said:
Thanks a lot for this. How about with a higher degree on the numerator - e.g. x^3 dx / (x^2 + x + 1)

$\noindent Use polynomial division first and the ``remainder term'' (remainder you get from doing the polynomial long division divided by the divisor, $x^2 + x +1$ here) will be amenable to the procedure outlined above. This is because the remainder has lower degree than the divisor. (The quotient term will just be a polynomial, which is easy to integrate.)$

Note that in your previous one where I added and subtracted something to get the numerator of lower degree than the denominator was actually just a shortcut method of polynomial division. (So even if the numerator has equal degree to the denominator, we need to divide first so that the numerator degree becomes strictly less than the denominator degree, after which we can use the earlier outlined method.)

Also note that if the denominator is a quadratic (without square root) with two easy roots and the numerator is a linear function (including constant function), we can also use the method of partial fractions relatively easily.

BenHowe · Feb 14, 2017

pikachu975 said:
How would you integrate x^2 dx / (x^2 - 2x + 3) lol sorry integration is so hard....

I got x + ln|x^2 - 2x+3| - (tan^-1 (x-1 / root2))/root 2 but it looks weird

I think you thought your question was $\int\frac{x^{2}}{\sqrt{x^2-2x+3}}dx$ not $\int\frac{x^{2}}{x^{2}-2x+3}dx$ cause of the sqrt on the denominator in the 1st q

Makes a substantial difference

pikachu975 · Feb 14, 2017

BenHowe said:
I think you thought your question was $\int\frac{x^{2}}{\sqrt{x^2-2x+3}}dx$ not $\int\frac{x^{2}}{x^{2}-2x+3}dx$ cause of the sqrt on the denominator in the 1st q

Makes a substantial difference

I did what integrand said and added and subtracted something from the numerator

InteGrand said:
$\noindent Use polynomial division first and the ``remainder term'' (remainder you get from doing the polynomial long division divided by the divisor, $x^2 + x +1$ here) will be amenable to the procedure outlined above. This is because the remainder has lower degree than the divisor. (The quotient term will just be a polynomial, which is easy to integrate.)$

Note than in your previous one where I added and subtracted something to get the numerator of lower degree than the denominator was actually just a shortcut method of polynomial division. (So even if the numerator has equal degree to the denominator, we need to divide first so that the numerator degree becomes strictly less than the denominator degree, after which we can use the earlier outlined method.)

Also note that if the denominator is a quadratic (without square root) with two easy roots and the numerator is a linear function (including constant function), we can also use the method of partial fractions relatively easily.

Thanks for the response but why do we use the remainder?

InteGrand · Feb 14, 2017

pikachu975 said:
I did what integrand said and added and subtracted something from the numerator

Thanks for the response but why do we use the remainder?

Its degree is less than the denominator's so we can use the method outlined earlier for that part.

pikachu975 · Feb 14, 2017

InteGrand said:
Its degree is less than the denominator's so we can use the method outlined earlier for that part.

So for example the remainder is x, then you use x/(x^2+x+1) ? Why does it still give the same answer?

InteGrand · Feb 14, 2017

pikachu975 said:
So for example the remainder is x, then you use x/(x^2+x+1) ? Why does it still give the same answer?

$\noindent You also have to include the quotient part (this part is easy to integrate as it is just a polynomial). So rewrite the fraction as a polynomial (quotient) plus the remainder divided by the dividend and go from there.$

InteGrand · Feb 14, 2017

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