## A community for students. Sign up today

Here's the question you clicked on:

## anonymous one year ago Integrate ((9sin^4(t) + 9cos^4(t))^1/2

• This Question is Closed
1. anonymous

It's all underneath the square root

2. Luigi0210

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

3. anonymous

Yes

4. 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?

5. anonymous

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

6. Luigi0210

You factored a square from each right?

7. Luigi0210

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

8. anonymous

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

9. anonymous

???

10. Luigi0210

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

11. anonymous

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

12. anonymous

??

13. 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. 14. Luigi0210 Yea that's what I was getting too 15. Luigi0210 And your num lock is off or something there? :P 16. anonymous I am doing arc length it's from 0 to pi/2 17. Luigi0210 That would of been a bit helpful to know 18. anonymous Answer in back says 3/2 19. EmmaTassone Maybe its a numerical solution or another way to calculate it, because you cant calculate it direclty with integrating methods. 20. IrishBoy123 you should post or scan the question it looks like an arc lngth, but why guess? :p 21. amistre64 might help to post the actual question ... or scan yeah 22. anonymous 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) 23. anonymous So found derivative and then did magnitude and got to where the question is now !! :) 24. anonymous :-/ 25. amistre64 what techniques have you covered? 26. 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 27. EmmaTassone i think you have derivated wrong, check it again 28. anonymous Cos^3(t) = -3sin^2(t) Sin^3(t) = 3cos^2(t) 29. amistre64 im thinking .. and it may not be any better y = sin^3(acos(cbrt(x))) to eliminate the parameter maybe? 30. EmmaTassone you are forgeting chain rule 31. amistre64 accursed chain rule lol 32. amistre64 writing it as say: (cos(t))^3 might make it easier to recall chaining it 33. anonymous Then what is it !? There is nothing with the t to do chain rule 34. amistre64 if u = cos(t) u^3 derives to 3u^2 u' 35. EmmaTassone \[(\cos ^{3}(x))\prime=3.\cos ^{2}(x).(-\sin(x))$

36. EmmaTassone

same with the sine

37. anonymous

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

38. anonymous

Gosh dang chain rule lol

39. EmmaTassone

yep, it was a calculation mistake lol

40. anonymous

Wow lol such stupid mistake lol

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy