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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.9
Evaluate the definite integrals in Exercises 176 to 195.
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Question : 184 of 271
Marks: +1, -0
∫01dx1−x2\int\limits_{0}^{1} \frac{dx}{\sqrt{1-x^2}}
Solution:  
∫01dx1−x2\int\limits_{0}^{1} \frac{dx}{\sqrt{1-x^2}}
=[sin⁡−1x]01=[\sin^{-1}x]^{1}_{0}
=sin⁡−1(1)−sin⁡−1(0)=\sin^{-1}(1)-\sin^{-1}(0)
=π2=\frac{\pi}{2}
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