Question

yoo! i'm stuck help guys! solve and explain all steps in the following questions : \[\int\limits_{}\sqrt{sinx}dx\]

Answer 1

-2 E(1/4 (pi-2 x)|2)+constant

Answer 2

yeah i bet you are stuck. i don't think this has a nice closed form

Answer 3

i took t=sqrt(sinx) so t^2 = sinx
then i did it as cosxdx for dt but at the end step i stucekd ,that it being unintegrable

Answer 4

forget it

Answer 5

e^(-st) it :)

Answer 6

euler would

Answer 7

:( wolframalpha answers, but really, explain me the steps of it?

Answer 8

if t^2 = sin(x); then 2t dt= cos(x) dx, and x = sin-1(t^2)
\[\int \frac{2t^2}{cos(sin^{-1}(t^2))}dt\]

not that that will be much easier, but if your gonna go that change of variable route it will turn to this i believe

Answer 9

wolfram is telling you nothing. it is telling you that the integral is an integral

Answer 10

well, i did the same thing, but after that part, i'm stuck...

Answer 11

\[\int \frac{2t^2}{\sqrt{(1-t^2)}}\ dt\]

Answer 12

hmm.... so now?

Answer 13

the 2t^2 should integrate by parts to 0 fairly quickly

Answer 14

or at least get you to a step that is easier :) or not

Answer 15

hmmm the triangle thing is really interesting and creative o.O

Answer 16

yeah, we need the cos of an angle whose sin was t^2 :)

Answer 17

its not able to do integrate by parts by the way... really did not deal a one like that before, its 2t^2 * dt / sqrt(1-t^2)

Answer 18

lets rewrite and see if it gets better

\[\int \frac{2t^2}{\sqrt{(1-t^2)}}\ dt\]
\[\int 2t^2*\frac{1}{\sqrt{(1-t^2)}}\ dt\]
\[=t *sin{-1}(t)-\int sin^{-1} (t)dt\]
maybe?

Answer 19

typoed it ... as usual

Answer 20

hmm looks like good to me, did not see a mistake there

Answer 21

\[=2t^2*sin^{-1}(t)-\int sin^{-1}(t)*4t\ dt\]

Answer 22

that sin^-1*

Answer 23

but here is the thing, how you defined cos(alfa) as equal to cos(sin^-1(t^2)) a trick on inverse functions?

Answer 24

yes, its using the meaning of it all;

cos(sin-1(t^2)) MEANS cos(angle), such that the sin(angle) = t^2

Answer 25

ahh k, have to work on trig funcs a bit more...

Answer 26

cos^2(a) + t^2 = 1

cos^2(a) = 1 - t^2

cos(a) = sqrt(1-t^2)

Answer 27

k now i got it that part thx

Answer 28

that parts a little deceptive since i did the math off

Answer 29

sin(a) = t^2 ; sin^2(a) = t^4

Answer 30

thats what i overlooked :)

Answer 31

its sin^-1(t^2) yea not sin-1(t^2)

Answer 32

:) :)

Answer 33

if you google this the answers are pretty funny. most are just plain wrong, some people say "none" but a good response is here
"So sorry, but this integral is not elementary.
In fact, we can reduce it to an elliptic integral.
To see this, let u = sin x, x = arcsin u,
dx = du/√(1-u²)
Then the integral becomes
∫√(u) du/√(1-u²),
and, rationalising the denominator,
∫ √(u-u³ du/(1-u²)

Since you have the square root of a cubic polynomial
in the integrand, you now have an elliptic integral."

Answer 34

it is not the case that just because you write down an integral you will be able to find some function in closed form whose derivative is the integrand.

Answer 35

hmm...

Answer 36

korcan never really does normal integrals :)

Answer 37

in fact the truth is that if you pulled a function out of a (rather large imaginary) hat the probability that it would be integrable is 0