Solving Integrals with Constants: Tips and Techniques for Success

[LagrangeEuler]

Homework Statement

How to solve integral
$$\int \frac{dy}{\sqrt{C_1-K \cos y-\frac{C_2}{\sin^2 y}}}$$

Homework Equations

$C_1,C_2$ and $K$ are constants.

The Attempt at a Solution

I am not sure which method I should use here or is this integral maybe eliptic? Please give me the hint. Which supstitution or method and I will solve integral to the end.[/B]

 

[micromass]

Start by eliminating the fractions under the square root.

 

[LagrangeEuler]

Start by eliminating the fractions under the square root.

Ok
$$C_1-K\cos y-\frac{C_2}{\sin^2 y}=\frac{C_1 \sin^2 y-K\cos y\sin^2 y-C_2}{\sin^2 y}$$
So now I have
$$\int \frac{\sin ydy}{\sqrt{C_1 \sin^2 y-K\cos y\sin^2 y-C_2}}$$

 

[LagrangeEuler]

Ok
$$C_1-K\cos y-\frac{C_2}{\sin^2 y}=\frac{C_1 \sin^2 y-K\cos y\sin^2 y-C_2}{\sin^2 y}$$
So now I have
$$\int \frac{\sin ydy}{\sqrt{C_1 \sin^2 y-K\cos y\sin^2 y-C_2}}$$

If I take
$\cos y =t$
then
$-\sin ydy=dt$
$$I=-\int\frac{dt}{\sqrt{C_1(1-t^2)-Kt(1-t^2)-C_2}}$$
$$I=-\int \frac{dt}{\sqrt{Kt^3-C_1t^2-Kt+C_1-C_2}}$$

 

[geoffrey159]

If your polynomial at the denominator were of degree 2, you would write it under its canonical form, and depending upon the coefficients and the discriminant, you would use a substitution of type Arcsin, Argsinh, or Argcosh to get rid of the square root. I don't know if you can do the same thing for polynomials of degree 3.

 

[Ray Vickson]

If I take
$\cos y =t$
then
$-\sin ydy=dt$
$$I=-\int\frac{dt}{\sqrt{C_1(1-t^2)-Kt(1-t^2)-C_2}}$$
$$I=-\int \frac{dt}{\sqrt{Kt^3-C_1t^2-Kt+C_1-C_2}}$$

You can express the integral in terms of Elliptic functions but it is very messy.

 

[RUber]

Are there limits on your integration?
It looks like the function of y might be defined for y in (0,pi)+k*pi for integer values of k, and may be imaginary for certain choices of constants.