\(\rm \int_0^1\frac{1}{\sqrt{4-x^2}}dx=\)

Question

Question

This question was previously asked in

Agniveer Vayu Group X 13 Oct 2023 Shift 1 Memory Based Paper

Answer 1

Concept:

\(\rm \int\frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}(\frac xa)+c\)

Calculation:

Let, I = \(\rm \int_0^1\frac{1}{\sqrt{4-x^2}}dx\)

= \(\rm \int_0^1\frac{1}{\sqrt{2^2-x^2}}dx\)

= \([\rm sin^{-1}(\frac x 2)]_0^1\) (∵ \(\rm \int\frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}(\frac xa)+c\))

= \(\rm sin^{-1}(\frac 1 2)-sin^{-1}(0)\)

= \(\frac\pi6\)

Hence, option (1) is correct.