Question
Download Solution PDF\(\rm \int_0^1\frac{1}{\sqrt{4-x^2}}dx=?\)
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
\(\rm \int\frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}(\frac xa)+c\)
गणना:
माना, I = \(\rm \int_0^1\frac{1}{\sqrt{4-x^2}}dx\)
= \(\rm \int_0^1\frac{1}{\sqrt{2^2-x^2}}dx\)
= \([\rm sin^{-1}(\frac x 2)]_0^1\) (∵ \(\rm \int\frac{1}{\sqrt{a^2-x^2}}dx=sin^{-1}(\frac xa)+c\) )
= \(\rm sin^{-1}(\frac 1 2)-sin^{-1}(0)\)
= \(\frac\pi6\)
इसलिए, विकल्प (1) सही है।
Last updated on Jun 30, 2025
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