Question

Hi everyone! Integrate from 0 to pi root[(2-2sin(theta))^2 +(-cos(theta))^2]...I posted a pic of the problem...thanks! :o)

Answer 1

ok this we can do !!

Answer 2

HOPEFULLY YOU CAN BREAK IT APART TO AVOID THE ROOT OF SINX! :O)

Answer 3

\[2-2\sin^2(x)+\cos^2(x)\]
\[=2(1-\sin^2(x))+\cos^2(x)\]
\[=2\cos^2(x)+\cos^2(x)=3\cos^2(x)\]

Answer 4

btw unless one of \(a\) or \(b\) is zero, it is never the case that
\[\sqrt{a+b}=\sqrt{a}+\sqrt{b}\] so you can never break it apart like that

Answer 5

I HAVE TO OPEN THIS WINDOW IN INTERNET EXPLORER...ALL YOUR FORMULAS ARE READING AS GIBBERISH...BRB

Answer 6

you are left with
\[\int\sqrt{3\cos^2(x)}dx\] or
\[\sqrt{3}|\cos(x)|dx\]

Answer 7

refresh the browser might work

Answer 8

satellite sir, how did you get 2-2 sin^2x ?

Answer 9

in the question it seems (2-2sin x)^2
which should become 4-8sin x+4sin^2 x

Answer 10

oh damn did i read it wrong?

Answer 11

\(\huge\int \limits_0^{\pi} \sqrt{(2-2\sin \theta)^2+(-\cos \theta)^2}\)

Answer 12

and dtheta

Answer 13

oh damn damn damn
this is probably going to be amazingly ugly

Answer 14

lets cheat and ask wolfram

Answer 15

wait!!!stop!!!super ...i made an error :o(

it's -2cos

Answer 16

wow!
http://www.wolframalpha.com/input/?i=sqrt%28%282-2sin%28x%29%29^2%2Bcos^2%28x%29%29

Answer 17

i so wished that there should be 2 there!!

Answer 18

now its probably easy

Answer 19

really sorry everyone...i'm seeing double right now and my computer is pausing every 10 seconds...place a 2 in front of cos

Answer 20

\(\huge \int \limits_0^{\pi}\sqrt{8-8\sin \theta}d\theta \)

Answer 21

\[(2-\sin(x))^2-2\cos^2(x)\]?

Answer 22

you can easily simplify that to 8-8 sin theta , can you ?

Answer 23

the plus minus part has got me stuck

Answer 24

oh lord is it
\[(2-\sin(x))^2+4\cos^2(x)\]??

Answer 25

yeah hartnn...i can get there...then pull a root 8 out and you are left with root 1-sintheta

Answer 26

@satellite73 sir, finally we are here,
\(\huge 2\sqrt2\int \limits_0^{\pi}\sqrt{1-\sin \theta}d\theta\)

Answer 27

then i think you apply some double angle formula or something to turn sin theta into cos(pi/2 - zero) i think but i can't remember exactly how

Answer 28

yeah i see it , i didn't understand it was
\[(2-\sin(x))^2+4\cos^2(x)\]

Answer 29

1- sin x = 1- sin 2 (x/2) = 1- 2 sin (x/2) cos (x/2) = sin^2 (x/2) -2 sin (x/2) cos (x/2) + cos^2 (x/2)
look at this carefully and tell me whether you understood the adjustments ?

Answer 30

is there a formula that says sin(theta) = cos(alpha - beta) or something like that?

Answer 31

lemme try and absorb what you wrote down

Answer 32

then,
\(\sqrt{1-\sin \theta }= \sqrt{(\sin (\theta/2)+\cos (\theta/2))^2}=\sin (\theta/2)+\cos (\theta/2)\)

Answer 33

wait, scratch that :P

Answer 34

why is 1- sin x = 1- sin 2 (x/2) ? I can't see how you got there

Answer 35

that step was correct, but the next step wasn't....
that was because, i wrote x = 2 (x/2)

Answer 36

hold on just a second...lemme show you what my professor did...

Answer 37

he turned 1-sin(theta) into 1 - cos(pi/2-0)

Answer 38

he said it was a double angle formula or something...then he multiplied the whole thing by a fancy form of "1" which was root2/root2

Answer 39

then he had 2root2root8 integral from 0 to pi of root[(1-cos(pi/2-0)/2

Answer 40

but this is where i get confused...not sure how he changed sin theta into cospi/2-0

Answer 41

did everyone leave me?

Answer 42

sorry, my PC hanged....
put u= sin theta

Answer 43

i won't go that(your profs) way...it lil bit confusing

Answer 44

u= sin theta, du=... ?

Answer 45

yeah...super confusing! i have no idea how sinx turn into cos(pi/2-0) ! yikes!

Answer 46

du is cos theta d theta...are you asking me or seeing if i know how to do it?

Answer 47

\(\sin x = \cos (\pi/2-x)\)
is an identity
i am trying to solve in an easier way

Answer 48

i see

Answer 49

crazy...we haven't integrated hard stuff for like 2 months and we are supposed to remember that obscure identity! geez! :o)

Answer 50

lol....i never remembered any formula....i used them so many times that they seem to be obvious to me...

Answer 51

i think i see what he did now when I am looking at my notes...he uses about 3 difficult techniques to get where he was going...not easy at all!

Answer 52

so, carrying on,
u = sin theta ----> cos theta = sqrt(1-sin^2 theta)= sqrt (1-u^2)
so,
du = cos theta dtheta gives
du = sqrt(1-u^2) dtheta
so that you can replace dtheta by
du/ sqrt(1-u^2)
got this ? its just algebraic manipulation...

Answer 53

any doubts in that ^ ?

Answer 54

i'm sifting through it right now...

Answer 55

i do not under stand this part cos theta = sqrt(1-sin^2 theta)...
i didn't have a cos theta in the integrand...it was root
(1-sintheta)...how did you get a cos ?

Answer 56

yes yes,
i put u= sin theta, right
but we needed du also
du = cos theta dtheta
so we will need cos theta in terms of u
so i used the identity, \(\sin^2u+\cos^2u=1\)

Answer 57

make that
\(\sin^2\theta +\cos^2\theta =1\)
\(u^2 +\cos^2\theta =1\)
\(\cos \theta = \sqrt{1-u^2}\)

Answer 58

no offense or anything but that puts a square root sign under another square root sign doesn't it? that seems even more confusing :o(

Answer 59

no it doesn't....i'll write out few steps, see whether u get it..

Answer 60

i will use x instead of theta for convenience.
\(\sqrt{1-\sin x}dx\)
u= sin x ---> \( \cos x = \sqrt {1-u^2} \)
\( du = \cos x dx = \sqrt {1-u^2} dx \)
so, dx= du / \( \sqrt {1-u^2} \)
so,
\(\sqrt{1-u}(du/ \sqrt {1-u^2} )\)
got this ? still confusing ?

Answer 61

lemme try and absorb what you wrote down

Answer 62

ok the part that is hanging me up is where you have \( \cos x = \sqrt {1-u^2} \)
shouldn't it be \( \cos "u" = \sqrt {1-u^2} \)...guess i don't see the transition from x to u

Answer 63

ok,
we had du = cos x dx
but we can use cos x in the integral, because we are changing the variable from x to u
so, i need cos x in terms of u
i have and i will use, u = sin x
so, \(u^2 = \sin^2 x\)
since i need cos x, i will write \(\sin^2 x = 1- \cos^2 x \)
\(u^2 =1- \cos^2 x\)
from that^ just isolate cos x

Answer 64

but we canNOT** use cos x

Answer 65

so cosx = root 1-u^2 right?

Answer 66

yes,
so u got till,
\(\huge \int \limits_0^{-1}\dfrac{\sqrt{1-u}}{\sqrt{1-u^2 }}du\)
note, i changed the limits, according to u= sin theta

Answer 67

now what?

Answer 68

next part is quite simple ....
\(\huge \int \limits_0^{-1}\dfrac{\sqrt{1-u}}{\sqrt{(1-u)\sqrt{(1+u)} }}du\)
because
1-u^2 = (1+u)(1-u)
note the \(\sqrt{1-u}\) part getting cancelled.

Answer 69

question...

Answer 70

root (1-u)/(1-u^2) = root (1/u) = root 1 / root u = 1/u^(1/2) = u^-(1/2) which then is integral of u^-(1/2) ? would that be correct?

Answer 71

root (1-u)/(1-u^2) = root (1/u) <---no
root (1-u)/(1-u^2) = root (1/{1+u})

Answer 72

ok...lemme write something real quick to see if i got it...

Answer 73

\(\huge \int \limits_0^{-1}\dfrac{1}{\sqrt{(1+u)} }du\)
this is quite an easy integral to solve...

Answer 74

did i do anything wrong there?

Answer 75

yes,
\((1-u^2)= (1+u)(1-u)\)
\((a^2-b^2)=(a+b)(a-b)\)

Answer 76

it isn't (1-u)^2
it actually is 1-u^2

Answer 77

ok...i think i get it now...however, whether i do it your way or my prof way, the darn problem is still very tricky!

Answer 78

yes, it seems so.
can you continue ?

Answer 79

so tell me...for the final answer, do you get 16-8root2 ?

Answer 80

nevermind...that was something else...thanks hartnn for all the help! :o)