anonymous
  • anonymous
Integrate ((9sin^4(t) + 9cos^4(t))^1/2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
It's all underneath the square root
Luigi0210
  • Luigi0210
This right? \(\Large \int (\sqrt{9sin^4 t +9cos^4 t}) dt\)
anonymous
  • anonymous
Yes

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Luigi0210
  • Luigi0210
I would trying reducing it down, factor out that 9 first: \(\Large \int \sqrt{9(sin^4 t +cos^4 t}) dt \) Since 9 can come out of the square roots, you get this: \(\Large \int 3(\sqrt{sin^4 t +cos^4 t}) dt \) Make sense so far?
anonymous
  • anonymous
Yep got that down already - then I tried to use half angle identities but got stuck there
Luigi0210
  • Luigi0210
You factored a square from each right?
Luigi0210
  • Luigi0210
\(\Large 3\int (\sqrt{(sin^2 t )^2+(cos^2 t)^2} dt \) ?
anonymous
  • anonymous
Yep then used 1+cos(2t)/2 and 1-cos(2t)/2
anonymous
  • anonymous
???
Luigi0210
  • Luigi0210
But yea, what did you get after you tried that?
anonymous
  • anonymous
Tried expanding them - but since its a square under square root can be factored out?
anonymous
  • anonymous
??
EmmaTassone
  • EmmaTassone
i dont want to be killjoy but i think that integral dont have solution. its similar to:\[\int\limits_{}^{}\sqrt{a.\cos^{2}(x)+b.\sin ^{2}(x)} And this integral dont have solution neither.Its called the elliptic integral.
Luigi0210
  • Luigi0210
Yea that's what I was getting too
Luigi0210
  • Luigi0210
And your num lock is off or something there? :P
anonymous
  • anonymous
I am doing arc length it's from 0 to pi/2
Luigi0210
  • Luigi0210
That would of been a bit helpful to know
anonymous
  • anonymous
Answer in back says 3/2
EmmaTassone
  • EmmaTassone
Maybe its a numerical solution or another way to calculate it, because you cant calculate it direclty with integrating methods.
IrishBoy123
  • IrishBoy123
you should post or scan the question it looks like an arc lngth, but why guess? :p
amistre64
  • amistre64
might help to post the actual question ... or scan yeah
anonymous
  • anonymous
Ok will try to write it out am on my iPad so here it goes Find the length of the following r(t) =
anonymous
  • anonymous
So found derivative and then did magnitude and got to where the question is now !! :)
anonymous
  • anonymous
:-/
amistre64
  • amistre64
what techniques have you covered?
anonymous
  • anonymous
I had thought to use the half angle formula to then take integral of ! In class covered take r'(t) then magnitude of r'(t) then take integral
EmmaTassone
  • EmmaTassone
i think you have derivated wrong, check it again
anonymous
  • anonymous
Cos^3(t) = -3sin^2(t) Sin^3(t) = 3cos^2(t)
amistre64
  • amistre64
im thinking .. and it may not be any better y = sin^3(acos(cbrt(x))) to eliminate the parameter maybe?
EmmaTassone
  • EmmaTassone
you are forgeting chain rule
amistre64
  • amistre64
accursed chain rule lol
amistre64
  • amistre64
writing it as say: (cos(t))^3 might make it easier to recall chaining it
anonymous
  • anonymous
Then what is it !? There is nothing with the t to do chain rule
amistre64
  • amistre64
if u = cos(t) u^3 derives to 3u^2 u'
EmmaTassone
  • EmmaTassone
\[(\cos ^{3}(x))\prime=3.\cos ^{2}(x).(-\sin(x))\]
EmmaTassone
  • EmmaTassone
same with the sine
anonymous
  • anonymous
So sine = 3sin^2(t)cos(t)
anonymous
  • anonymous
Gosh dang chain rule lol
EmmaTassone
  • EmmaTassone
yep, it was a calculation mistake lol
anonymous
  • anonymous
Wow lol such stupid mistake lol

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