Question

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

Answer 1

It's all underneath the square root

Answer 2

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

Answer 3

Yes

Answer 4

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?

Answer 5

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

Answer 6

You factored a square from each right?

Answer 7

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

Answer 8

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

Answer 9

???

Answer 10

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

Answer 11

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

Answer 12

??

Answer 13

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.

Answer 14

Yea that's what I was getting too

Answer 15

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

Answer 16

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

Answer 17

That would of been a bit helpful to know

Answer 18

Answer in back says 3/2

Answer 19

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

Answer 20

you should post or scan the question


it looks like an arc lngth, but why guess?

:p

Answer 21

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

Answer 22

Ok will try to write it out am on my iPad so here it goes
Find the length of the following
r(t) = <cos^3(t), sin^3(t)

Answer 23

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

Answer 24

:-/

Answer 25

what techniques have you covered?

Answer 26

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

Answer 27

i think you have derivated wrong, check it again

Answer 28

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

Answer 29

im thinking .. and it may not be any better

y = sin^3(acos(cbrt(x))) to eliminate the parameter maybe?

Answer 30

you are forgeting chain rule

Answer 31

accursed chain rule lol

Answer 32

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

Answer 33

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

Answer 34

if u = cos(t)

u^3 derives to 3u^2 u'

Answer 35

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

Answer 36

same with the sine

Answer 37

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

Answer 38

Gosh dang chain rule lol

Answer 39

yep, it was a calculation mistake lol

Answer 40

Wow lol such stupid mistake lol