Question

Evaluate the definite integrals: $\int^1_0 {\frac{dx}{\sqrt{1-x^2}}$

Answer 1

Answer:

Step-by-step explanation:

Formula used:

$\int{\frac{1}{\sqrt{1-x^2}}}\:dx=sin^{-1}x+c$

Now,

$Now,\\\\\int\limits^{1}_0{\frac{1}{\sqrt{1-x^2}}}\:dx\\\\=[sin^{-1}x]^{1}_0\\\\=sin^{-1}(1)-sin^{-1}(0)\\\\=\frac{\pi}{2}$