# Thread: An integral with trigonometric functions

1. ## An integral with trigonometric functions

Hello,

I'm trying to solve the following integral:

$\int \frac{\cos x}{\sqrt{\sin 2x}} dx$

Here's what I've tried:

$\int \frac{\cos x}{\sqrt{\sin 2x}} dx=$

$\int \frac{\cos x}{\sqrt{2 \sin x \cos x}} dx=$

$\frac{1}{\sqrt{2}} \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx=$

universal substitution; $t=\tan \frac{x}{2}$; $\sin x=\frac{2t}{1+t^2}$; $\cos x=\frac{1-t^2}{1+t^2}$; $dx=\frac{2dt}{1+t^2}$

...and then I get

$\int \sqrt{\frac{1-t^2}{t}} \cdot \frac{dt}{1+t^2}$

Can you give me a clue how to tackle this integral?

Thanks!

2. ## This May be helpful

I don't think this has an antiderivatitive, but It does look alot like the Beta function or the Incomplete Beta function.

Beta function - Wikipedia, the free encyclopedia

Do you have limits of integration or do you need an antiderivative?

3. Originally Posted by gusztav
Hello,

I'm trying to solve the following integral:

$\int \frac{\cos x}{\sqrt{\sin 2x}} dx$

Can you give me a clue how to tackle this integral?

Thanks!
Here's what I'd do: Yeah, I'd try some things first but if I couldn't figure it out, I'd input it into Mathematica so that I could at least figure out how to arrive at the answer:

$\int \frac{\cos(x)}{\sqrt{\sin(2x)}}dx=-\frac{cos(x)\sqrt{\sin(2x)}}{3 \sqrt{\sin(x)}}\sum_{k=0}^{\infty}\frac{(3/4)_k (3/4)_k}{(7/4)_k}\frac{\cos^{2k}(x)}{k!}$

At this point I'd ask myself, why is Mathematica reporting this? Can it be that there is no finite antiderivative? Mathematica is sometimes wrong of course and frequently expresses answers in "uncommon" ways. If there is no finite answer, then I'd try and figure out how did Mathematica come up with that answer (which is in terms of a hypergeometric series)?

4. Originally Posted by shawsend
Here's what I'd do: Yeah, I'd try some things first but if I couldn't figure it out, I'd input it into Mathematica so that I could at least figure out how to arrive at the answer:

$\int \frac{\cos(x)}{\sqrt{\sin(2x)}}dx=-\frac{cos(x)\sqrt{\sin(2x)}}{3 \sqrt{\sin(x)}}\sum_{k=0}^{\infty}\frac{(3/4)_k (3/4)_k}{(7/4)_k}\frac{\cos^{2k}(x)}{k!}$

At this point I'd ask myself, why is Mathematica reporting this? Can it be that there is no finite antiderivative? Mathematica is sometimes wrong of course and frequently expresses answers in "uncommon" ways. If there is no finite answer, then I'd try and figure out how did Mathematica come up with that answer (which is in terms of a hypergeometric series)?
What I find amazing is that I can differentiate that expression in Mathematica and it returns the integrand. How is it doing that? I just don't understand big business.

5. $\frac{1}{2}\int \frac{2\cos x}{\sqrt{\sin 2x}}dx$

$=\frac{1}{2}\int\frac{\cos x+\sin x+\cos x-\sin x}{\sqrt{\sin 2x}}dx$

$=\frac{1}{2}\int\frac{\cos x+\sin x}{\sqrt{1-(\sin x-\cos x)^2}}dx$+ $\frac{1}{2}\int\frac{\cos x-\sin x}{\sqrt{(\sin x+\cos x)^2-1}}dx$

Now,put $t=\sin x-\cos x$ and $u=\sin x+\cos x$

$=\frac{1}{2}\int\frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int\frac{du}{\sqrt{u^2-1}}$

$=\frac{1}{2}sin^{-1}t+\frac{1}{2}ln|u+\sqrt{u^2-1}|+c$

$=\frac{1}{2}sin^{-1}(\sin x-\cos x)+\frac{1}{2}ln|(\sin x+\cos x)+\sqrt{\sin 2x}|+c$

6. Can ${\sqrt {2\tan x}} +c$ be the solution?

$\frac{1}{\sqrt{2}} \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx$

$\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt {\tan x}}$

$\frac{1}{\sqrt{2}} \int ({\tan x})^{- \frac{1}{2}}$

$\frac{2(\tan x)^{\frac {1}{2}}}{\sqrt{2}}$

7. that's wrong, check the basics of integration.

8. Originally Posted by chengbin
Can ${\sqrt {2\tan x}} +c$ be the solution?

$\frac{1}{\sqrt{2}} \int \frac{\sqrt{\cos x}}{\sqrt{\sin x}} dx$

$\frac{1}{\sqrt{2}} \int \frac{1}{\sqrt {\tan x}}$

$\frac{1}{\sqrt{2}} \int ({\tan x})^{- \frac{1}{2}}$

$\frac{2(\tan x)^{\frac {1}{2}}}{\sqrt{2}}$
You could make the substiution $u=tan(x)$, but then you have

$\frac {1}{\sqrt{2}} \int u^{- \frac {1}{2}}cos^{2}(x)du$