The value of $\int_0^1\int_0^1\frac{dx dy}{\sqrt{(1-x²)(1-y²)}}$ is
- A. $\frac{\pi²}{4}$
- B. $\frac{\pi}{4}$
- C. $\frac{\pi²}{2}$
- D. $\frac{\pi}{2}$
Answer: A. $\frac{\pi²}{4}$
Explanation: $I = \int_0^1\int_0^1\frac{dx dy}{\sqrt{(1-x²)(1-y²)}} = (\int_0^1\frac{dx}{\sqrt{1-x²}})(\int_0^1\frac{dy}{\sqrt{1-y²}})$. Since $\int_0^1\frac{dx}{\sqrt{1-x²}} = [\sin^{-1}x]_0^1 = \sin^{-1}(1) = \frac{\pi}{2}$. $I = \frac{\pi}{2} \times \frac{\pi}{2} = \frac{\pi²}{4}$.