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Please Solve R.D.Sharma class 12 Chapter 19 Definite Integrals Exercise 19.1 Question 3 Maths textbook Solution.

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Answer: \frac{\pi }{6}

Hint:: Use indefinite formula to solve the integral and then put the value of limit to get the required answer

Given: \int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^{2}}}dx


\int_{0}^{\frac{1}{2}}\frac{1}{\sqrt{1-x^{2}}}dx=\left [ \sin ^{-1}x \right ]^{\frac{1}{2}}_{0}                                                                                \left [ \int \frac{1}{\sqrt{1-x^{2}}}dx=\sin ^{-1}x \right ]

=\sin ^{-1}\left(\frac{1}{2}\right)-\sin ^{-1}(0) \quad\left[\sin \left(\frac{\pi}{6}\right)=\frac{1}{2} \Rightarrow \sin ^{-1}\left(\frac{1}{2}\right)=\frac{\pi}{6}, \sin 0=0 \Rightarrow \sin ^{-1}(0)=0\right]

=\left [ \frac{\pi }{6}-0 \right ]=\frac{\pi }{6}

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