How do you integrate xdx/sqrt(x-x^2) (exercise 4.3 question 1h) from terry lee)?
Thanks for the help
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// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
Last edited by Drongoski; 14 Feb 2017 at 2:25 PM.
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// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
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
1st Year BAppFinBActStud @ MQ
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
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]
1st Year BAppFinBActStud @ MQ
1st Year BAppFinBActStud @ MQ
Using my personal way of integrating, I note that the derivative of x-x^{2} 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.
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Last edited by pikachu975; 14 Feb 2017 at 2:54 PM.
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
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.
Last edited by InteGrand; 14 Feb 2017 at 5:14 PM.
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
2016 HSC (Accelerated)
// 2 Unit Maths // Studies of Religion 1 //
2017 HSC
// Biology // Physics // Maths Extension 1 // Maths Extension 2 // English Advanced //
Last edited by InteGrand; 14 Feb 2017 at 4:57 PM.
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
1st Year BAppFinBActStud @ MQ
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