Could someone please help me solve this integral: I = squareroot(1-x^2) dx
Start by substituting x = sin(w), so that dx = cos(w) dw. You'll end up having to integrate $\displaystyle \cos^2(x)dx$, and to do that you can use the half angle formula: $\displaystyle \cos^2w = \frac 1 2 (1+\cos(2w))$. Try it, and post back with what you get.
You may want to check this page which has detailed instructions(and demonstration) on how to integrate $\displaystyle \int \sqrt{1 - x^2}\,\,dx$ as a definite integral. For indefinite integral the procedure is same.
definite integrals
To make it easier I re-edited the post. We have to evaluate:
$\displaystyle \int \sqrt{1-x^2}\,\,dx$
Here it is again:
Let's think of $\displaystyle x$ and $\displaystyle 1$ in the above question as two sides of a right triangle.
$\displaystyle \cos{(\theta)} = \frac{x}{1} = x$
$\displaystyle \sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}$
$\displaystyle \text{And } dx = -\sin{(\theta)}\,\,d\theta$
$\displaystyle \begin{align*}\int \sqrt{1-x^2}\,\,dx =& -\int \sin^2{(\theta)}\,\,d\theta \\=& \int \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2} + C \\=& \frac{\sin{(\theta)}\cos{(\theta)}}{2} - \frac{\theta}{2} + C......\text{[using trigonometry]}\\=& \frac{x\sqrt{1-x^2}}{2} - \frac{\cos^{-1}{(x)}}{2}+ C\\& \text{.....[by replacing } \sin{(\theta)}, \cos{(\theta)} \text{ and } \theta \text{]}\end{align*}$
Hope it helps.