Evaluate the definite integral int_{0}^{1} fr | vedclass

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(B) Let $I = \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$.We know that the antiderivative of $\frac{1}{\sqrt{1-x^{2}}}$ is $\sin^{-1} x$.Let $F(x) = \sin^

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$\frac{\pi}{2}$

Vedclass · Answer

(B) Let $I = \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$.We know that the antiderivative of $\frac{1}{\sqrt{1-x^{2}}}$ is $\sin^{-1} x$.Let $F(x) = \sin^{-1} x$.By the second fundamental theorem of calculus,we have $I = F(1) - F(0)$.Substituting the limits,we get $I = \sin^{-1}(1) - \sin^{-1}(0)$.Since $\sin^{-1}(1) = \frac{\pi}{2}$ and $\sin^{-1}(0) = 0$,we have $I = \frac{\pi}{2} - 0$.Therefore,$I = \frac{\pi}{2}$.

Evaluate the definite integral $\int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$.

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