Integration by splitting the numerator (1 Viewer)

pikachu975 · Feb 14, 2017

Integral of (5-x^2)dx / sqrt(3-2x-x^2)

By splitting the numerator, thanks for the help!

InteGrand · Feb 14, 2017

pikachu975 said:
Integral of (5-x^2)dx / sqrt(3-2x-x^2)

By splitting the numerator, thanks for the help!

$\noindent One way is as follows (general method when the numerator is a quadratic and denominator is square root of a quadratic). Say we want to find $\int \frac{q(x)}{\sqrt{Q(x)}}\,\mathrm{d}x$, where $Q(x)$ is a degree $2$ polynomial and $q(x)$ is as well.$

$\noindent 1. Write $q(x) = aQ(x) + bQ'(x) + c$ for some wise choice of constants $a,b,c$ (again you can try and show as an exercise that such constants will always exist (and in fact will be unique)).$

$\noindent 2. Observe that this causes the integral to become$

$a\int \sqrt{Q(x)}\,\mathrm{d}x + b\int\frac{Q'(x)}{\sqrt{Q(x)}}\,\mathrm{d}x + c\int \frac{1}{\sqrt{Q(x)}}\,\mathrm{d}x.$

$\noindent $\Big{(}$For the first integral we used that $\frac{Q(x)}{\sqrt{Q(x)}} = \sqrt{Q(x)}$.$\Big{)}$$

$\noindent 3. Evaluate each integral. (Each of these integrals has a standard method of solving and you are expected to be able to do each of them.)$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent One way is as follows (general method when the numerator is a quadratic and denominator is square root of a quadratic). Say we want to find $\int \frac{q(x)}{\sqrt{Q(x)}}\,\mathrm{d}x$, where $Q(x)$ is a degree $2$ polynomial and $q(x)$ is as well.$

$\noindent 1. Write $q(x) = aQ(x) + bQ'(x) + c$ for some wise choice of constants $a,b,c$ (again you can try and show as an exercise that such constants will always exist (and in fact will be unique)).$

$\noindent 2. Observe that this causes the integral to become$

$a\int \sqrt{Q(x)}\,\mathrm{d}x + b\int\frac{Q'(x)}{\sqrt{Q(x)}}\,\mathrm{d}x + c\int \frac{1}{\sqrt{Q(x)}}\,\mathrm{d}x.$

$\noindent $\Big{(}$For the first integral we used that $\frac{Q(x)}{\sqrt{Q(x)}} = \sqrt{Q(x)}$.$\Big{)}$$

$\noindent 3. Evaluate each integral. (Each of these integrals has a standard method of solving and you are expected to be able to do each of them.)$

Thanks again for the help, can you please show me how to write the (5-x^2) in the form you stated, because I don't quite understand it. Thanks!

InteGrand · Feb 14, 2017

pikachu975 said:
Thanks again for the help, can you please show me how to write the (5-x^2) in the form you stated, because I don't quite understand it. Thanks!

$\noindent In your particular example it is $q(x) = 5-x^{2}$ and $Q(x) = 3-2x -x^{2}$, so $Q'(x) = -2-2x$. So try finding constants $a,b,c$ such that$

$$5-x^{2}\equiv a\left(3-2x -x^{2}\right) + b(-2-2x) + c$.$

$\noindent (You can find these similarly to how you find constants in partial fractions decomposition when it's written out like that. Or since this system is ``triangular'', you can find the constants easily by first equating $x^{2}$ coefficients, then $x$, then constant terms. (Don't worry if you don't know what triangular means here.))$

pikachu975 · Feb 14, 2017

InteGrand said:
$\noindent In your particular example it is $q(x) = 5-x^{2}$ and $Q(x) = 3-2x -x^{2}$, so $Q'(x) = -2-2x$. So try finding constants $a,b,c$ such that$

$$5-x^{2}\equiv a\left(3-2x -x^{2}\right) + b(-2-2x) + c$.$

$\noindent (You can find these similarly to how you find constants in partial fractions decomposition when it's written out like that. Or since this system is ``triangular'', you can find the constants easily by first equating $x^{2}$ coefficients, then $x$, then constant terms. (Don't worry if you don't know what triangular means here.))$

I tried it and got a = 1, b = -1, c = 0, but I don't think it's right

5-x^2 = -ax^2 - 2x (a+b) + (3a-2b+c)
So a = 1 by equating coefficients
a+b = 0, so b = -1
3a-2b+c = 5
3 + 2 + c = 5
c = 0

Did I do something wrong?

InteGrand · Feb 14, 2017

pikachu975 said:
I tried it and got a = 1, b = -1, c = 0, but I don't think it's right

5-x^2 = -ax^2 - 2x (a+b) + (3a-2b+c)
So a = 1 by equating coefficients
a+b = 0, so b = -1
3a-2b+c = 5
3 + 2 + c = 5
c = 0

Did I do something wrong?

$\noindent You did it right. You can (and should) check your answer, i.e. check that $5-x^{2} =\left(3-2x -x^{2}\right) -(-2-2x)$.$

BenHowe · Feb 14, 2017

Hey, here's my solution. It took me a while tho...

$\int\frac{5-x^{2}}{\sqrt{3-2x-x^{2}}}dx\\=-\int\frac{x^2}{\sqrt{3-2x-x^{2}}}dx+5\int\frac{1}{\sqrt{3-2x-x^{2}}}dx\\now\hspace{0.1cm}3-2x-x^{2}=-(x^{2}+2x-3)\\=-((x+1)^{2}-4)\\=4-(x+1)^{2}\\I=-\int\frac{x^{2}}{\sqrt{4-(x+1)^{2}}}dx+5\int\frac{1}{\sqrt{4-(x+1)^{2}}}dx\\Let\hspace{0.1cm}J=-\int\frac{x^{2}}{\sqrt{4-(x+1)^{2}}}dx\\let\hspace{0.1cm}x+1=2sin\theta\\dx=2cos\theta d\theta\\J=-\int\frac{x^{2}cos\theta}{cos\theta}d\theta\\=-\int(2sin\theta-1)^{2}d\theta\\=-\int4sin^{2}\theta-4sin\theta cos\theta+1\hspace{0.1cm}d\theta\\=-\int3-2cos2\theta-2sin2\theta\hspace{0.1cm}d\theta\\=-[3\theta-sin2\theta+cos2\theta]+c\\=-3sin^{-1}(\frac{x+1}{2})+\frac{(x+1)\sqrt{3-2x-x^{2}}}{2}-1+\frac{(x+1)^{2}}{2}+c\\Let\hspace{0.1cm} K=5\int\frac{1}{\sqrt{4-(x+1)^{2}}}dx\\=5sin^{-1}(\frac{x+1}{2})+c\\I=J+K\\=2sin^{-1}(\frac{x+1}{2})+\frac{(x-3)\sqrt{3-2x-x^{2}}}{2}+c$

Note 10 attempts at editing it to get the equation to be valid... rip

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Integration by splitting the numerator (1 Viewer)

pikachu975

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InteGrand

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pikachu975

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InteGrand

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InteGrand

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BenHowe

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