Question

what is the integral of square root of sin x?

Answer 1

Is there some bound in particular you want to evaluate it over, like \[\int_0^{\pi/2} \sqrt{\sin x} dx\]

Answer 2

e^ix=cosx+isinx, can we use this for sqrts too

Answer 3

ah nvm

Answer 4

I was thinking about trying to square a power series with unknown coeficients and set that equal to sin x so that I could solve for the coefficients that way

Answer 5

oo interesting

Answer 6

I made some progress I'll show, but at least we could get a power series representaion for it :P

Answer 7

lemme see

Answer 8

theres a laplace form or fourier infinite series form

Answer 9

to findiing the coefficients

Answer 10

not a big fan of that though, the coefficients are just as messy

Answer 11

Ok so first off:

\[\sin x = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2n+1} = \sum_{n=0}^\infty a_nx^n= \]

The problem I had is that this sorta like skips even numbers, so finding a taylor series of this form that squares to it I think will be difficult.
\[\left( \sum_{k=0}^\infty b_k x^k \right)^2\]

Although the convolution is pretty straight forward (and can probably be simplified too.
\[a_n = \sum_{k=0}^n b_kb_{n-k}\]

Remember, \(a_{2n}=0\), anyways I think I have a better way though, since squaring to an odd sequence of terms is quite messy, I have a slightly altered approach!

Answer 12

yeah gotcha

Answer 13

Let's instead factor out an x, because:

\[\left(\sum_{k=0}^\infty b_{2k+1} x^{2k+1} \right)^2 =\sum_{k=0}^\infty a_{2k} x^{2k} \]

But we want something of the form:


\[\left(\sqrt{x} \sum_{k=0}^\infty b_{2k+1} x^{2k+1} \right)^2 =\sum_{k=0}^\infty a_{2k+1} x^{2k+1} = \sin x\]

Which I think might end up having a nicer and easy to find closed form.

Answer 14

Take \(\arcsin u=x\), so that \(u=\sin x\) and \(\dfrac{\mathrm{d}u}{\sqrt{1-u^2}}=\mathrm{d}x\).
\[\int_0^{\pi/2}\sqrt{\sin x}\,\mathrm{d}x=\int_0^1\frac{\sqrt u}{\sqrt{1-u^2}}\,\mathrm{d}u\]
Then setting \(u=\sqrt t\), so that \(u^2=t\) and \(\dfrac{\mathrm{d}t}{2\sqrt t}=\mathrm{d}u\), you get
\[\int_0^1\frac{t^{1/4}}{\sqrt{1-t}}\cdot\frac{\mathrm{d}t}{2\sqrt t}=\frac{1}{2}\int_0^1 t^{-1/2}(1-t)^{-1/4}\,\mathrm{d}t\]which has a nice closed form with the face of the Euler beta function.
https://en.wikipedia.org/wiki/Beta_function

Answer 15

From sine: \[a_{k} = \frac{(-1)^k}{(2k+1)!}\]

Power series:

\[\sqrt{\sin x} = \sqrt{x} \sum_{n=0}^\infty b_{n}x^{2n}\]

Convolution:
\[a_{n} = \sum_{k=0}^{n} b_{k}b_{n-k}\]
\[\frac{(-1)^k}{(2k+1)!} = \sum_{k=0}^{n} b_{k}b_{n-k}\]

This splits up into two cases when n is even or odd, although I don't really see a clean way out from here so I'm gonna stop here haha.

Answer 16

Here's a fun identity that I looked at in trying to solve it:

\[i \int_a^b \sqrt{\sin x} dx = \int_{a+\pi}^{b+\pi} \sqrt{\sin x} dx \]

Ultimately though, I think the problem is solved by @SithsAndGiggles I guess if closed forms are what you're looking for.