Ideas for this integral: $\int \frac{\sqrt{\tan{x}}}{\sin{x}} dx$

Question

$$\int \frac{\sqrt{\tan{x}}}{\sin{x}} \mathrm{d}x$$

So I was wondering if this correctly by converting $\sqrt{\tan{x}}$ into $\frac{\sqrt{\sin{x}}}{\sqrt{\cos{x}}}$ therefore I can divide it with $\sin{x}$ and that will give me $\frac{\sqrt{\cos{x}}}{\sqrt{\sin{x}}}$ and that will be $\sqrt{\cot{x}}$ so the integral is formed into $\displaystyle \int \sqrt{\cot x}\mathrm{d}x $. Is this so far correctly done?

No it has not been simplified correctly. $\frac{\sqrt{\tan x}}{\sin x} = \frac 1{\sqrt{\sin x \cos x}} = \sqrt 2 \cdot \sqrt {\csc 2x}$. — Deepak, Aug 05 '20 at 16:46
Can you elaborate on how you can divide the fraction by $\sin x$? I don't understand. — an4s, Aug 05 '20 at 16:48
This is a special integral (incomplete elliptic integral of the first kind): — Anindya Prithvi, Aug 05 '20 at 16:49
mistakes happen but dont delete this question, i want to see someone do it — Anindya Prithvi, Aug 05 '20 at 16:49
@AnindyaPrithvi Yes, it's the incomplete elliptic integral of the first kind, well said! — euler_med, Aug 05 '20 at 16:51
@EM4 are you sure this is an indefinite integral, i.e., there are no limits? — an4s, Aug 05 '20 at 16:53
I made this problem up in my head, i got excited by $\int \sqrt{tan{x}} dx$ and I was like hmmm can this happen. Until I seen my error and @AnindyaPrithvi stated. — EM4, Aug 05 '20 at 16:55
This is not exactly a duplicate since your integral is indefinite but might still be of interest to you. — an4s, Aug 05 '20 at 16:57
$-\sqrt{\sin (2 x)} \sqrt{\tan (x)} \csc (x) F\left(\left.\frac{\pi }{4}-x\right|2\right)$ where F is https://en.wikipedia.org/wiki/Elliptic_integral#Incomplete_elliptic_integral_of_the_first_kind — flinty, Aug 05 '20 at 17:21

score 5 · Answer 1 · answered Aug 05 '20 at 17:34

5

$$I=\int \frac{\sqrt{\tan x}}{\sin x} dx$$ Let $\tan x =t^2 \implies \sec^2 x dx=2t dt$ Then $$I=\int \frac{2}{\sqrt{1+t^4}} dt$$ This can be found interms of Elliptic functions.

answered Aug 05 '20 at 17:34

Z Ahmed

46,319

B&F 263.50: $I=F(2\tan^{-1}\sqrt{\tan x},m=1/2)$ – Parcly Taxel Dec 11 '20 at 17:29

score 1 · Answer 2 · answered Aug 09 '20 at 22:07

1

Hint:

$$\int \frac{\sqrt{\tan{x}}}{\sin{x}} dx=\int\frac{1}{\sqrt{\cos\left(x\right)\sin\left(x\right)}} dx=\int\frac{\sqrt{2}}{\sqrt{2}\sqrt{\cos\left(x\right)\sin\left(x\right)}}dx=\int\frac{\sqrt{2}}{\sqrt{\sin\left(2x\right)}}dx$$

Substituting $t=2x$ you have $dx=\frac 12dt$:

$$=\frac{1}{2}\int\frac{dt}{\sqrt{\sin\left(t\right)}}$$

where the last integral can be reduced to a complete elliptical integral of first species.

answered Aug 09 '20 at 22:07

Sebastiano

8,290

where you found the 2x? – EM4 Aug 10 '20 at 13:37
1

I dived for $\sqrt2$ numerator and denominator. Put $\sqrt2$ into $\sqrt{\cos(x)\sin(x)}$, i.e. $\sqrt{2\sin(x)\cos(x)}=\sqrt{\sin(2x)}$. Hence $2x=t$. – Sebastiano Aug 10 '20 at 18:49

Ideas for this integral: $\int \frac{\sqrt{\tan{x}}}{\sin{x}} dx$

2 Answers2

Hint: