Trigonometric Integration: Solving ∫ dx/√(1-k*sin2x)

[msandeep92]

Hi everyone,

In my research project, i am struck with an integration. Can someone help me out:

∫ dx/√(1-k*sin²x)

Thanks,
Sandeep

 

[Infinitum]

Hey Sandeep.

How did you attempt this integral? It will be easier(and ethical) to help you out if you share how you worked on it.

 

[msandeep92]

I tried with substitution √k*sinx = t;

But it ends up with ∫dt/√[(1 - t²/k)*(1 - t²)]

I don't know how to go ahead with this.

I tried with ksin²x = t also. But that too doesn't work.

Please help me. This is one of the last parts of a huge integration i am doing...

Thanks,
Sandeep.

 

[Infinitum]

Try breaking $1-ksin^{2}x$ into $(1-\sqrt{k}sin x)(1+\sqrt{k}sinx)$

Now try splitting the term into two simple integrals of the form $\frac{1}{a+bsinx}$

 

[msandeep92]

But it is √(1 - ksin²x).

So, how can we split it into two simple integrals?

Sandeep.

 

[Infinitum]

Oh, I misread, and did not take into consideration the square root. Apologies.

As for the function ∫ dx/√(1-k*sin2x), there is no simple way to integrate it. It is concerned with Elliptical integrals : http://mathworld.wolfram.com/EllipticIntegraloftheFirstKind.html
http://en.wikipedia.org/wiki/Elliptic_integral

 

[msandeep92]

Thank you.

Will Go through it..

Sandeep.