FP2 Calculus Watch

davidcy147
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#1
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#1
Can anybody help me integrate (x+2)/(4-x^2)^1/2 ?
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Notnek
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#2
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Split up the fraction and then use a substitution, u=4-x^2 for the first fraction and recognize the integral of the second fraction.
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lordcrusade9
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well split it as x/(4-x^2)^1/2 and 2/(4-x^2)^1/2

the former is reverse of chain rule. the latter is arcsin function.
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thelostchild
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#4
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split it up like so
$\int \frac{x}{\sqrt{4-x^2}}dx+\int \frac{2}{\sqrt{4-x^2}}

for the first one its a simple substitution, and the second one you should recognize the form for a trig sub
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davidcy147
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#5
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#5
Can you explain the reverse of the chain rule part please?
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thelostchild
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#6
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its just integration by substitution you should remember this from C4 (if you do edexcel)
we define u=4-x^2
therefore
du=-2x dx

dx=\frac{du}{-2x}

we substitute them into the integral like so

$\int \frac{x}{-2x\sqrt{u}}du

the the x's cancel so were left with

$\int-\frac{1}{2\sqrt{u}}du

remember it now?
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davidcy147
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#7
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#7
By doing that I end up with -4(4-x^2)^1/2 which isnt right
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thelostchild
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#8
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you integrated it wrong, its a C1 integral you really need more practice at these can rewrite that as:
$\int-\frac{1}{2}u^{-0.5}dx

now remember the rule
$\int ax^n dx=\frac{ax^{n+1}}{n+1}
where a and n are constants

I get as the answer

-\sqrt{4-x^2}+c
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