Let's think of $\displaystyle x$ and $\displaystyle 1$ in the given question as two sides of a triangle like in this image below:

Attachment 28076
$\displaystyle \cos{(\theta)} = \frac{x}{1} = x$

$\displaystyle \sqrt{1-x^2} = \sqrt{1-\cos^2{(\theta)}} = \sin{(\theta)}$

$\displaystyle \text{And } dx = -\sin{(\theta)}\,\,d\theta$

Now for lower limit $\displaystyle x = 0$:

$\displaystyle x = \cos{(\theta)} = 0 \therefore \theta = \frac{\pi}{2}$

And for upper limit $\displaystyle x = 1$:

$\displaystyle x = \cos{(\theta)} = 1 \therefore\,\, \theta = 0$

Back to the main question which can be written as:

$\displaystyle \begin{align*}\int_0^1 \sqrt{1-x^2}\,\,dx =& -\int_{\frac{\pi}{2}}^0 \sin^2{(\theta)}\,\,d\theta \\=& \int_{\frac{\pi}{2}}^0 \frac{\cos{(2\theta)}-1}{2}\,\,d\theta....\text{[Using trigonometry]} \\=& \frac{\sin{(2\theta)}}{4} -\frac{\theta}{2}\Bigg]_{\frac{\pi}{2}}^0 \\=& 0 - 0 - \frac{\sin{(\pi)}}{4} + \frac{\pi}{4} \\=& \frac{\pi}{4}....\text{[for 1st quadrant]}\end{align*}$