Can you solve without modifying the integrand? (My twelfth integration problem)

You can write the equation of half a circle using the formula $y = \sqrt{1-x^{2}}$ . $\dfrac{dy}{dx} = \dfrac{-x}{\sqrt{1-x^{2}}}$ . The formula for arc length states that $s = \sqrt{1 + \left( \dfrac{dy}{dx} \right)^{2}}$ . Notice that the integral given in this problem is an arc length integral. It's essentially asking, what's a fourth of the circumference of a circle? Well, the circumference of a whole unit circle as given in the problem is $2 \pi$ , so then divide that by four, and the answer becomes $\dfrac{\pi }{2 }$ , and since $a= 1, b=2$ , the answer is $1+2 = \boxed{3}$ .

$\int_{0}^{1} \sqrt{1 + \frac{x^2}{1 - x^2}} dx = \int_{0}^{1} \frac {1}{\sqrt{1 - x^2}} dx = arcsin(1) - arcsin (0) = \frac{\pi}{2}$

Substitute x for $\large\sin \theta$ . The integral simplifies to a very simple form.