1. ## Integration

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

Thanks for the help

2. ## Re: Integration

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

3. ## Re: Integration

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

4. ## Re: Integration

Hey I know it's been answered but I still wanna upload it so I can ask a question. pikachu integral image.png. How can I put the latex in the actual post like the others above? I only know how to link as a picture

5. ## Re: Integration

Originally Posted by BenHowe
Hey I know it's been answered but I still wanna upload it so I can ask a question. pikachu integral image.png. 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

6. ## Re: Integration

Ok lemme try [tex] $\int\frac{x}{\sqrt{x-x^{2}}}dx$

$=-\frac{1}{2}\int\frac{-2x}{\sqrt{-x^{2}+x}}dx$

$=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{-x^{2}+x}}dx$

$now -x^{2}+x=-(x^{2}-x)$

$=-((x-\frac{1}{2})^{2}-\frac{1}{4})$

$=(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}$

$=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}}}dx$

$=-\sqrt{-x^{2}+x}+\frac{1}{2}sin^{-1}(2x-1)+c\hspace{0.1cm}$ [\tex]

7. ## Re: Integration

Originally Posted by BenHowe
Ok lemme try [tex] $\int\frac{x}{\sqrt{x-x^{2}}}dx$

$=-\frac{1}{2}\int\frac{-2x}{\sqrt{-x^{2}+x}}dx$

$=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{-x^{2}+x}}dx$

$now -x^{2}+x=-(x^{2}-x)$

$=-((x-\frac{1}{2})^{2}-\frac{1}{4})$

$=(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}$

$=-\frac{1}{2}\int\frac{-2x+1}{\sqrt{-x^{2}+x}}-\frac{1}{\sqrt{(\frac{1}{2})^{2}-(x-\frac{1}{2})^{2}}}dx$

$=-\sqrt{-x^{2}+x}+\frac{1}{2}sin^{-1}(2x-1)+c\hspace{0.1cm}$ [\tex]
You don't press enter for a new line maybe it is \\

8. ## Re: Integration

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

9. ## Re: Integration

Originally Posted by pikachu975
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.

10. ## Re: Integration

Originally Posted by Drongoski
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

11. ## Re: Integration

Originally Posted by pikachu975
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
$\noindent Steps to evaluating any integral like this (linear function on numerator and degree two polynomial (quadratic) in denominator or in a square root in denominator) is as follows. Let q(x) be the quadratic present (it was x -x^{2} in your case).$

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

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

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

12. ## Re: Integration

Originally Posted by pikachu975
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
$\noindent Add and subtract -2x +3 to the numerator of the integrand, so the integral becomes \int \left(1 + \frac{2x-3}{x^{2} -2x +3}\right)\,\mathrm{d}x = x + \int \frac{2x-3}{x^2 -2x +3}\,\mathrm{d}x. This latter integral can be tackled using the procedure outlined above.$

13. ## Re: Integration

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

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

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

$\noindent 3. Evaluate these two integrals. The first can be done by inspection or substituting u = q(x), \mathrm{d}u = q'(x)\,\mathrm{d}x. The second can be done e.g. by completing the square to make it of the form of a standard integral (though it may not be on the Standard Integrals sheet anymore).$
Thanks a lot for this. How about with a higher degree on the numerator - e.g. x^3 dx / (x^2 + x + 1)

14. ## Re: Integration

Originally Posted by pikachu975
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.

15. ## Re: Integration

Originally Posted by pikachu975
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

16. ## Re: Integration

Originally Posted by BenHowe
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

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

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?

17. ## Re: Integration

Originally Posted by pikachu975
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.

18. ## Re: Integration

Originally Posted by InteGrand
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?

19. ## Re: Integration

Originally Posted by pikachu975
So for example the remainder is x, then you use x/(x^2+x+1) ? Why does it still give the same answer?
$\noindent You also have to include the quotient part (this part is easy to integrate as it is just a polynomial). So rewrite the fraction as a polynomial (quotient) plus the remainder divided by the dividend and go from there.$

20. ## Re: Integration

$\noindent E.g. By polynomial division, \frac{x^{3}}{x^{2} + 1} = \underbrace{x}_{\text{quotient}} - \underbrace{\frac{x}{x^{2}+1}}_{\text{remainder part''}}, so$

\begin{align*}\int \frac{x^{3}}{x^{2} +1}\,\mathrm{d}x &= \int \left( x -\frac{x}{x^{2}+1}\right)\,\mathrm{d}x \\ &= \cdots .\end{align*}

21. ## Re: Integration

Originally Posted by InteGrand
$\noindent E.g. By polynomial division, \frac{x^{3}}{x^{2} + 1} = \underbrace{x}_{\text{quotient}} - \underbrace{\frac{x}{x^{2}+1}}_{\text{remainder part''}}, so$

\begin{align*}\int \frac{x^{3}}{x^{2} +1}\,\mathrm{d}x &= \int \left( x -\frac{x}{x^{2}+1}\right)\,\mathrm{d}x \\ &= \cdots .\end{align*}
Thank you for the help as well as Drongoski and BenHowe

22. ## Re: Integration

Originally Posted by InteGrand
$\noindent E.g. By polynomial division, \frac{x^{3}}{x^{2} + 1} = \underbrace{x}_{\text{quotient}} - \underbrace{\frac{x}{x^{2}+1}}_{\text{remainder part''}}, so$

\begin{align*}\int \frac{x^{3}}{x^{2} +1}\,\mathrm{d}x &= \int \left( x -\frac{x}{x^{2}+1}\right)\,\mathrm{d}x \\ &= \cdots .\end{align*}
How did you write the quotient and remainder part underneath the terms, it looks so epic

23. ## Re: Integration

Pikachu the only reason I said that you mighta thought it had the square root in the denominator because originally I thought it did as well and got the same answer as you

24. ## Re: Integration

Originally Posted by BenHowe
How did you write the quotient and remainder part underneath the terms, it looks so epic
I used

Code:
\underbrace{...}_{\text{...}}
.

(Replace the first "..." with the thing you want the brace to go under and the second "..." with what text you want to write under the brace.)

25. ## Re: Integration

Originally Posted by BenHowe
Pikachu the only reason I said that you mighta thought it had the square root in the denominator because originally I thought it did as well and got the same answer as you
I checked I didn't have it, I used integrand's method stated somewhere above

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