Question on integration

Hey

How do I integrate the following function?

..........1
A = 2π∫[(3+√(1-x^2))*√(1 + (x / √(1-x^2))^2]dx
.........-1

The interval is -1 to 1.

Thanks

You can simplify that last radical expression significantly.

√[1 + (x/√(1 - x²))²] = √[(1 - x² + x²)/(1 - x²)] = 1/√(1 - x²).

So your integrand is actually

3/√(1 - x²) + 1.

The integrand is even, and the interval is symmetric, so integrate from 0 to 1 and double the result. You'll get an inverse sine function.

1
∫ 4π (3/√(1 - x²) + 1) dx = 12π arcsin(1) + 4π - 0 = 6π² + 4π.
0

Looks like some sort of surface area.

If I read the equation correltly, then,

A = 2*pi int_{-1}^{1} [(3+sqrt(1-x^2))* sqrt(1+[x^2/(1-x^2)]) dx
= 2*pi int_{-1}^{1} [(3+sqrt(1-x^2)) /[sqrt(1-x^2)] dx
= 2*pi int_{-1}^{1} [(3/[sqrt(1-x^2)] dx + 2*pi int_{-1}^{1} 1 dx
Now compute each integration sepratly,
2*pi int_{-1}^{1} 1 dx = 4*pi.

For the first part (2*pi int_{-1}^{1} [(3/[sqrt(1-x^2)] dx ), set x=cos(y) , then dx = sin(y) dy, when x=-1, y= pi, and when x=1, y=2*pi. Not also 1-cos^2 (y) = sin^2 (y). Therefore,

2*pi int_{-1}^{1} [(3/[sqrt(1-x^2)] dx = 6*pi int_{pi}^{2*pi} sin(y)/sin(y) dy
= 6*pi int_{pi}^{2*pi} 1 dy = 6*(pi^2).

Thus A = 6*(pi^2) + 4*pi = 71.7841.

Best Regards.