INTEGRATION

We do it by trigonemetric substitution.

Let x = sinp. So dx/dp = cosp & dx = cosp.dp
1 - x^2 = 1 - sin^2(p) = cos^2(p)

Now substitute everything back into our original integral, giving:
INT(cos^2(p).dp)
We use the trig identity:
cos^2(p) = (1-cos2p)/2
I'm sure you can integrat that. When you're done, you find your answer in terms of p. But notice p = arcsin(x), so you substitute that back in.

Why did we make that seemingly arbitrary substitution in the beginning? Well, firstly, substitution is one of the oldest and most common tricks in maths. Also, we note that the trig substitution gets rid of the nasty square root, and then we can evaluate a simpler expression.

well thank you very much....that just opened a whole new universe for me..

Are you doing Calc 2? You should have done trig substitutions in Calc 1. What textbook are you using? Stewart's "Calculus and Concepts" is a good revision book on Calculus, although maybe not the best absolute best, if you want to be a real whizz at calc.

yes i just started cal2 and no we didnt do any trig substitution in cal1 (or we might have but i didnt know we did). We are using "Calculus of a Single Variable - (8th edition)" by Larson, Hostetler and Edwards