Mathematics
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OpenStudy (anonymous):
integral of dy/sqrt(4-y^2)?
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OpenStudy (freckles):
a trig sub will work just dandy here
OpenStudy (inkyvoyd):
you'll want to complete the square and what you're left with use trig sub
OpenStudy (anonymous):
\[\int\limits \frac{ dy }{ \sqrt{4-y ^{2}}}\]
OpenStudy (freckles):
Recall
1-sin^2(x)=cos^2(x)
so
4-4sin^2(x)=4cos^2(x)
so 4-(2sin(x))^2=4cos^2(x))
OpenStudy (freckles):
so it should be obvious what you want to replace y with
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OpenStudy (anonymous):
i am not good with trig identities
OpenStudy (freckles):
that's fine
i wrote the identity you needed
OpenStudy (freckles):
when is 4-(2sin(x))^2 the same as 4-y^2
when y= what?
OpenStudy (anonymous):
oh.. i see
y=2sinx
OpenStudy (freckles):
so that is your sub
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OpenStudy (anonymous):
ok thank you!
OpenStudy (freckles):
@inkyvoyd I'm interested in the partial fractions way
OpenStudy (freckles):
but how do you do it with the sqrt around the 4-y^2 part/
OpenStudy (inkyvoyd):
@freckles ... you don't. I am such an idiot.
OpenStudy (freckles):
oh
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OpenStudy (anonymous):
wait freckles what happens after i sub in 2sinx
OpenStudy (anonymous):
what happens to sqrt
OpenStudy (freckles):
y=2sin(x)
dy=?
OpenStudy (freckles):
And inky I don't believe you are an idiot. :p
OpenStudy (anonymous):
dy=2cosx
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OpenStudy (inkyvoyd):
that's the problem @freckles ... that's how idiots get to the top of important positions :P
OpenStudy (freckles):
dy=2cos(x) dx
OpenStudy (freckles):
So replace dy with 2cos(x) dx
replace y with 2sin(x)
OpenStudy (freckles):
and use the identity tht 4-4sin^2(x)=4cos^2(x)