- anonymous

Integrate ((9sin^4(t) + 9cos^4(t))^1/2

- schrodinger

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- anonymous

It's all underneath the square root

- Luigi0210

This right? \(\Large \int (\sqrt{9sin^4 t +9cos^4 t}) dt\)

- anonymous

Yes

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## More answers

- 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

Yep got that down already - then I tried to use half angle identities but got stuck there

- Luigi0210

You factored a square from each right?

- Luigi0210

\(\Large 3\int (\sqrt{(sin^2 t )^2+(cos^2 t)^2} dt \) ?

- anonymous

Yep then used 1+cos(2t)/2 and 1-cos(2t)/2

- anonymous

???

- Luigi0210

But yea, what did you get after you tried that?

- anonymous

Tried expanding them - but since its a square under square root can be factored out?

- anonymous

??

- 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

Yea that's what I was getting too

- Luigi0210

And your num lock is off or something there? :P

- anonymous

I am doing arc length it's from 0 to pi/2

- Luigi0210

That would of been a bit helpful to know

- anonymous

Answer in back says 3/2

- EmmaTassone

Maybe its a numerical solution or another way to calculate it, because you cant calculate it direclty with integrating methods.

- IrishBoy123

you should post or scan the question
it looks like an arc lngth, but why guess?
:p

- amistre64

might help to post the actual question ... or scan yeah

- 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

So found derivative and then did magnitude and got to where the question is now !! :)

- anonymous

:-/

- amistre64

what techniques have you covered?

- 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

i think you have derivated wrong, check it again

- anonymous

Cos^3(t) = -3sin^2(t)
Sin^3(t) = 3cos^2(t)

- amistre64

im thinking .. and it may not be any better
y = sin^3(acos(cbrt(x))) to eliminate the parameter maybe?

- EmmaTassone

you are forgeting chain rule

- amistre64

accursed chain rule lol

- amistre64

writing it as say: (cos(t))^3 might make it easier to recall chaining it

- anonymous

Then what is it !? There is nothing with the t to do chain rule

- amistre64

if u = cos(t)
u^3 derives to 3u^2 u'

- EmmaTassone

\[(\cos ^{3}(x))\prime=3.\cos ^{2}(x).(-\sin(x))\]

- EmmaTassone

same with the sine

- anonymous

So sine = 3sin^2(t)cos(t)

- anonymous

Gosh dang chain rule lol

- EmmaTassone

yep, it was a calculation mistake lol

- anonymous

Wow lol such stupid mistake lol

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