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  • Evaluate the definite integrals in Exercises 1 to 20. Integral 0 to 1 dx/root of 1 - x^2

Evaluate the definite integrals in Exercises 1 to 20.

    Q9.    \int_0^1\frac{dx}{\sqrt{1-x^2}}

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Given integral: \int_0^1\frac{dx}{\sqrt{1-x^2}}

Consider the integral \int \frac{dx}{\sqrt{1-x^2}}

\int \frac{dx}{\sqrt{1-x^2}} = \sin^{-1}x

So, we have the function of x, f(x) = \sin^{-1}x

Now, by Second fundamental theorem of calculus, we have

I = f(1) -f(0)

= \sin^{-1}(1) -\sin^{-1}(0)

= \frac{\pi}{2} - 0

= \frac{\pi}{2}

Posted by

Divya Prakash Singh

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